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V  ", 

THE  THEORY 


OF 


THE  CONTINUOUS  GIRDER 


Its  Applicatioq  to  Girders  with  and   without  Variable 
Cross-sections, 


BY 


A.    HOWE,    C. 

Professor  of  Civil  Engineering,  Rose  Polytechnic  Institute. 


UNIVERSITY 


NE\V     YORK: 

Eqgiqeering    News   Publishing   Co, 

1889. 


Entered  according  to  an  Act  of  Congress,  1888,  by  MALVERD  A.  HOWE,  in  the  office  of  the  Librarian  of  Congres 

at  Washington. 


MOOKK  &  IjANGfEN, 

Printers, 
Terre  Haute,  Ind, 


PREFACE. 


Continuous  bridges — with  the  exception  of  draw  bridges- 
are  not  considered  economical  and  are  not  designed  by. 
American  engineers.  This  probably  accounts  for  the  brief 
treatment  the  theory  of  the  continuous  girder  receives  in 
text  books  and  engineering  literature. 

With  one  or  two  exceptions,  all  American  treatises  con- 
sider the  moment  of  inertia  as  constant  and  deduce  two 
equations  for  the  moment  over  any  support,  one  to  be  ap- 
plied when  the  loads  are  on  the  left  of  the  support  and  the 
other  when  the  loads  are  on  the  right.  These  two  equa- 
tions combined  would,  of  course,  give  the  moment  over  any 
support  for  any  load,  but  only  when  the  moment  of  inertia 
could  be  assumed  as  constant,  as  in  girders  with  parallel 
flanges,  where  it  might  be  undue  refinement  to  consider  the 
cross-section  as  variable. 

American  engineers  have,  until  recently,  assumed  the 
moment  of  inertia  as  constant  even  when  the  flanges  of  the 
girder  were  considerably  inclined,  as  in  the  Sabula  draw 
bridge,  relying  upon  the  factor  of  ignorance  to  cover  all  dis- 
crepancies which  might  arise  from  the  assumption.  A  few 
years  ago  the  modulous  of  elasticity  was  quite  variable  in 
large  bridges — though  all  engineers  considered  it  as  constant 
in  computations — but  now,  since  experience  shows  that  iron 
and  steel  can  be  manufactured  with,  practically,  a  constant 
modulus  of  elasticity,  it  may  safely  be  considered  as  con- 
stant, without  leading  to  any  appreciable  error. 

Although  no  material  may  be  saved  by  considering  the 


IV  PRP:FACE. 

moment  of  inertia  as  variable  yet,  if  it  is  so  considered, 
the  material  will  be  placed  where  it  will  do  the  best  ser- 
vice. 

Girders  with  inclined  chords,  computed  and  designed  as 
if  their  cross-sections  were  constant,  have  more  material 
than  is  necessary  in  some  of  the  compression  pieces,  and 
not  enough  in  some  of  the  tension  members,  as  is  clearly 
shown  by  the  results  on  page  77  of  the  text. 

The  object  of  the  following  pages  is  to  present  to  com- 
puters, engineers  and  students  a  complete  mathematical 
treatment  of  the  theory  of  the  continuous  girder,  and  show  how 
it  can  be  applied  to  any  girder — especially  to  fixed  girders 
and  girders  of  two  and  three  spans — under  any  conditions. 

With  the  single  assumption  that  the  modulous  of  elas- 
ticity is  constant  a  general  equation  has  been  deduced  for 
the  moment  over  any  support  of  any  girder  under  any 
conditions  of  loading,  of  any  length  of  spans,  of  variable 
or  constant  cross-section  and  for  supports  at  any  level. — By 
difference  of  level  of  the  supports  is  meant  any  change  of 
level  which  may  take  place  when  the  girder  is  in  position. 
It  is  evident  that  such  a  change  alone  would  affect  the  mo- 
ments.— From  this  equation  special  equations  are  readily 
deduced  for  any  particular  case.  To  illustrate  the  sim- 
plicity of  the  transformations  necessary  for  any  special  case 
and  also  for  the  convenience  of  engineers,  equations  have 
been  given  for  all  the  special  cases  usually  discussed  in  text 
books,  and  also  equations  for  these  cases  when  the  moment 
of  inertia  is  considered  as  variable. 

The  usual  general  equations  for  reactions,  deflections  and 
intermediate  moments  and  their  transformed  equations  for 
special  cases  are  also  presented. 

Several  examples  are  solved  to  illustrate  the  application 
of  the  formulas  and  to  show  that  the  processes  are  almost 
mechanical  when  the  formulas  are  thoroughly  understood. 
From  those  examples  which  are  solved  considering  the  mo- 
ment of  inertia  as  constant  and  then  variable,  a  good  idea 
of  the  manner  in  which  the  moment  of  inertia  affects  the 


PKEFACE.  V 

results  can  be  obtained.     In  the  case  of  the   Sabula  draw 
there  is  a  difference  of  about  twenty  per  cent. 

In  all  pin  connected  bridges  the  loads  are  considered  to 
be  concentrated  at  the  apices  or  panel  points.  In  com- 
puting the  moments  for  such  girders  Table  I.  will  be  found 
to  be  very  convenient,  as  in  it  are  found  expressions  for 

rfk-tW+k"  and  A- — k'1  for  all  values  of  k=  /     from   0.001    to 
0.999  inclusive. 

The  works  enumerated  under  References,  page  107,  have 
been  consulted,  and  some  of  their  parts  used  without  any 
material  change,  for  which  credit  is  given  in  foot  notes. 

The  author  is  indebted  to  R.  H.  Brown,  C.  E.,  and  Geo. 
H.  Hutchinson,  C.  E.,  for  valuable  assistance  and  sugges- 
tions. 

M.  A.  H. 

TERRE  HAUTE,  IND.,  September,  1888. 


GENERAL  CONTENTS. 


PACiE. 

Nomenclature 1-2 

I. 

GENERAL    RELATIONS. 

Conditions  of  equilibrium  In  any  girder 4 

General  relations  betweenthe  moments  and  reactions  and  the  loads  .  .  4-6 
Equation  for  the  moment  over  any  support  when  the  moment  of  inertia 

is  variable 7 

Equation  for  the  moment  over  any  support  when  the  momen  1  of  inertia 

is  constant 12 

General  equations  for  Shear 15-16 

General  equations  for  Deflection .  ...  17 

II. 

SUPPORTED  GIRDERS. 

I.  A  simple  girder  resting  upon  two  supports 19-20 

II.  A  beam  continuous  over  three  supports ....  21-26 

III.  A  beam  continuous  over  four  supports  .    .    .  27-28 

IV.  "The  Tipper" 29-83 

III. 

BEAMS  WITH  FIXED  ENDS. 

I.  A  beam  fixed  at  one  end,  and  supported  at  the  othf1' 34-38 

II.  A  beam  fixed  at  both  ends     i(9-41 

III.  A  beam  fixed  at  one  end,  and  unsupported  at  the  othei     42-43 

IV.  A  beam  011  two  supports,  and  one  end  unsupported 44 

V.  A  beam  on  one*support,  having  one  end  fixed,  and  the  other  unsup- 

ported        45 

VI.  A  beam  on  two  supports,  having  neither  end  supported 6 

IV. 
THE  POINT  OF  ZERO  MOMENT. 

General  equations  with  graphical  determination  of  the  point  of  zero 

moment    .  .  47-51 


vni  GENERAL  CONTENTS. 

V. 

APPLICATIONS. 

PAGE. 
Examples  illustrating  the  application  of  the  formulas o2-XM 

APPENDIX. 

Determination  of  the  equation  of  the  elastic  line 8<>-87 

Demonstration  of  Equation  (A),  p.  7 89-100 

Demonstration  of  the  equation  of  the  elastic  line 80-88 

General  expression  for  the  Theorem  of  Three  Moments &7 

Demonstration  of  Equation  (A) ' 8W-10O 

TABLE  I. 

Values  for  k— 2  k*+k3  and  k— k*  for  all  ratios  ~  =  k  from  .mil  to  .9W*,  in- 
clusive     101-UHi 

REFERENCES. 

Some  references  to  monographs,  periodicals,  Ac.,  which  consider  the 

theory  of  the  continuous  girder 1(>7-108 

INDEX. 

Index  of  Equations      .  XKMlo 

si  MMARV. 

Summary  of  contents 111-11(1 


NOMENCLATURE. 


number  of  the  support  just  at  the  left  of  the  /•"  span. 
r=The  horizontal  length  of  the  span  r. 
-Pr=Any  concentrated  load  in  the  i**  span. 
tc€r=Any  uniform  load,  per  lineal  foot,  in  the  j**  span. 
af=The  distance  from  the  left  support  r  to  any  concentrated 

load  Pr.     a=J:rlr, 
€tr'=The  distance  from  the  support  »•  to  the  point  where  the 

uniform  load  ends  in  the  /•"  span. 
O/'— The  distance  from  the  left  support  r  to  the  point  where 

the  uniform  load  begins  in  the  /•**  span. 

T  flr 

Kj=  -T-,  or,  ar=kr  lr , 

xr=The  distance  from  the  left  support  r  to  any  point  in  the 

Jw*  span. 

JIr==rhe  bending  moment  over  the  support  r. 
bending  moment  over  any  support  m. 

bending  moment  at  any  section,  xr  from  the  sup- 
port r. 

JJT=The  bending  moment  at  the  center  of  any  span  of  a 
girder. 

shear  just  at  the  right  of  the  support  /'. 

shear  just  at  the  left  of  the  support  r. 

reaction  at  the  support  r,  and  equals  Sr^-Sr'. 

distance  the  support  )'  is  below  some  specified  hori- 
zontal line  of  reference. 
n  of  summation. 


2  THE  CONTINUOUS  GIRDER. 

£,.=Tangent  of  the  angle  that  the  elastic  line  makes  with  the 
horizontal  over  the  support  r. 

yr=The  deflection  of  1he  girder  at  any  section  or,. ;  y/,.is  meas- 
ured from  the  horizonal  line  of  reference. 

S— The  number  of  spans. 


e,— The  distance  from  the  left  support  r  to  the  point  where 
the  moment  of  inertia  of  the  section  of  the  girder 
changes  for  the  fi^t  time  in  the  span  i*. 

e2=The  distance  to  the  point  where  it  changes  the  second 
time. 

ev=The  distance  to  any  point  where  the  moment  of  inertia 
changes,  always  measured  from  the  left  support  •/•  and 
in  the  span  r. 

Ii=The  moment  of  inertia  between  the  last  value  of  ee  and 
the  end  of  the  •/*"*  span.  See  Fig. 

_t>=The  moment  of  inertia  between  the  support  r  and  the 
pointy. 

j,— The  moment  of  inertia  between  et  and  e2  in  the  r("  span. 

,7,.— The  moment  of  inertia  at  any  section  in  the  rm  span. 
12=The  modulus  of  elasticity. 


GENERAL  RELATIONS. 


*  In  the  '/**  span  of  a  continuous  girder,  whose  length  is 
£r,  Fig.  1,  take  a  point  o  vertically  above  the  r'"  support, 
as  the  origin  of  co-ordinates,  and  the  horizontal  through  o 
as  the  axis  of  abscissas.  Suppose  any  section  of  the  girder 
at  a  distance  wr  from  the  left  support  r,  and  between  this 
section  and  the  support  r  a  weight  _P,  distant  from  r, 
Or=kr  /,.. 

Now,  if  the  girder  is  continuous  over  any  number  of  sup- 
ports there  will  be  a  bending  moment  over  each  of  them  (if 
the  ends  are  not  fixed  the  bending  moment  over  the  first 
and  last  support  will  equal  zero):  JC-/  over  r — 1,  lMLr  over  r, 
Hfr+i  over  r-\-lj  etc.,  also,  there  will  be  a  shear  Sr  just  at  the 
right  and  $,.  just  at  the  left  of  /*,  $.+,,  just  at  the  right  and 
Sr+i  just  at  the  left  of  r-\-l,  etc.  If  there  is  equilibrum  the 
following  conditions  must  be  fulfilled  : 

*   See  "Continuirlichen  Und  Einfachen  Trager,"  page 4,  by  Prof.  Weyraucb. 
"Theory  and  Calculation  of  Continuous  Bridges,"  page  52,  by  Prof.  Mer- 
riman. 

"Strains  in  Framed  Structures,"  page  244,  by  Prof.  DuBois. 


4  THE  CONTINUOUS  GIRDER. 

/.      The  algebraic  sum  of  all  the  horizontal  forces  must  be  zero. 
II.     The  algebraic  sum  of  all  the  vertical  forces  must  be  zero. 
III.      The  algebraic  sum  of  the  moments  of  all  the  forces  must 
be  zero. 

From  the  third  condition  we  have  for  any  section  in  the 
r'"  span, 

M,.     Mr-4-Sr  x—  P,.  (xr—  a,'.)    ....    x£ar     .    .    ...    .    .    .(1) 

If  in  (1)  we  make  w,.~ln  .Mr~Mr+l,  and  it  becomes 
Mr+i=-M,A  -Sf  I,.  —  P,  (lr  —  a,.),  or,  since  a,     &,.  /,.,  we  have 

Mr+i  =  M,A  -Sr     lr—Pr     I     (1  —  ^          ......  .       .    (2) 

From  (2), 


And  if  there  is  no  load  in  the  span  r  this  becomes 

r=M.±L-Mr 
t 

^,—P,—Sn  therefore  from  (3), 
^  -K-M«L.:pf  t,  ,«,.  better, 


And  if  there  is  no  load  in  the  span  '/*  —  /,  (o)  l)ecomes 


The  reaction  at  any  support  equals  the  sum  of  the  shears 
at  that  support,  or 


GENERAL  RELATIONS.  5 

The  above  formulas  were  deduced  under  the  supposition 
that  there  was  but  a  single  concentrated  load  JP,  in  the  span 
r.  If  there  be  more  than  one  concentration,  the  formulas 
become  : 

(1)  Mr^=Mf-\  ••$,.  x,.—l'  Pr  (x,.--a,.)  .    .    .  x£a,  ......  (8) 

(2)  M^^lC+flt  lr-£  P,  /,  (/-*)     .........  W 

(3)-  N,-  -^±Mf      ^P,  (/-  A-(.)    ..........  (10) 


partial  uniform    load  itt  span  r.  —  If  to, 

represents  the  uniform  load  per  lineal  foot  in  the  span  /•, 
we  can  write 


,.,  fc,.  /,. 


I  ft=     «,,  da,,  or,  since  a,.,  fc 


,.  /,  d  k,.=lwf  I,.  kf)        £       (12) 

ar=kflf  fc'-V 

The  last  expression  in  (12)  indicates  the  difference  of  the 
values  of  the  parenthesis  when  A',  equals  — -  and  — -  re- 
spectively. 

Substituting  (12)  in  the  above  equations,  we  have, 
From  (8), 

Mr=Mr+Srxr--Cwr  /,  d  kr  (x—kf  I,}  .    .    .  x£a, (13) 

<fv 


6  THE  CONTINUOUS  GIRDER. 

From  (9), 

a,. 
Mr+l^Mr-S,.  l-w,  l*ffd  k,.  (7-ik,) (14) 

a, 

Or 


Mr+i^Mr-\  -8,  l—wr  l*r  (k  —  4-  tf)  :. (15) 


From  (10), 


From  (11), 

M_       ]f  j  ^ 


Uniform  load  over  entire  apan  r.—lf  the  entire 
span  is  uniformly  loaded,  we  have  a,.-~ln  d'f--o  and  a,f—\  I,.. 

From  (13),     (In  this  case  a,---xr,  a,.  =o  and  af — $  x,  since 
the  load  considered  cannot  extend  beyond  jr,.). 

Mr=M,-{  -»Sfr  air  —  -^-  w,  r; (18) 

From  (15), 


GENERAL  RELATIONS.  7 

From  (16), 

4^Vf*<  ......    .......    ..(20) 

From  (17), 


These  twenty-  one  general  equations  apply  to  all  spans 
and  conditions  of  loading,  but  before  they  can  be  solved,  it 
will  be  necessary  to  find  expressions  for  Jf^or  Sin  terms  of 
the  loads,  lengths  of  the  spans,  etc. 

This  can  be  done  by  introducing  the  equation  of  the 
elastic  line  and  the  theorem  of  three  moments.  The  following 
expression  is  obtained  for  the  bending  moment  over  any 
support  m',  E  alone,  being  constant. 


A 


In  which  the  respective  expressions  have  the  following 
values : 


*  A  complete  demonstration  of  this  equation  is  given  in  the  Appendix 
and  should  be  thoroughly  understood  before  any  attempt  is  made  to  apply 
equation  A  practically. 


8  THE  CONTINUOUS  GIRDER. 

A, 

Con  cent  rated  loads— 

A,=  —2  P,  /;  V  Av-  ->  A1;     #)  *U  •    •        •  See  (57)   ....</) 

Partial  uniform  load  — 

From  (12), 


-  Pr--wr  I,,   (k,.)  therefore  (i)  becomes 


a',=kf  /, 

»,->  ............  (74) 


Uniform  load  orer  all— 


Uniform  load  over  all— 

A,=  -  w,.  Ill  <>,._, 

4 


B, 

Concentra  ted  loads— 

B,.=  -S  P,  I-  (Av-A:;)  or.  .......  See  (58)  .        .    ,  (./) 

Partial  uniform  toads— 


GENERAL  RELATIONS.  9 

Concentrated  loads- 
Partial  uniform  load— 

ar=kr!r 
"r  =    -Fr  *,  /;  or_,  |  fe,-  -L  A-f  |  +Hr  or_t  .    ,  (78) 

Uniform  load  over  all— 

Concentrated  loads— 

'^=  ~2F;_,  IPr_s !,_,  (l—k,.^  O-ILL,  Or...  See  (59)  . . .  (//.) 
Partial  uniform  load— 


;L,=  -2  F;_,  o.w^i^,  kr_t-  ~v_,  [    -Hr'_<or  .  .  (so) 


Uniform  load  over  all— 

;L<=  -  F;L,  or  w^  I^-H^  or  .  (si) 

Ht 

Concentrated  loads— 


V=/r 


10  THE  CONTINUOUS  GIRDER. 

Partial  -uniform  load— 

Substitute  the  following  expressions  in  (/)  : 

or 
I  Pr  (rv—a,y  =  Cwf  d  a,,  (e—ajf  ....  a'r<e,    ....  (82) 

«y 


,- 

?  Pr  (ev—ar)s=  Cwr  d  a,.  (c0—ar)s  .    .    -  ..  dr<ev   .....  (83) 
a'f 

Probably  the  easiest  way  to  handle  this  case  is  to  consider 
the  uniform  load  as  extending  from  the  left  support  to  the 
farther  end  of  the  load  and  integrate  between  the  limits 
ar=ev  and  d'r=o;  then  consider  the  load  as  extending  from 
the  left  support  to  the  nearer  end  of  the  load  and  integrate 
between  the  limits  a',=d^ev  and  o,  and  take  the  difference 
of  the  results. 

Uniform  load  over  all— 

dar(ev-ary=~wr  £  .    .  .    .(84) 


t.  y 

IPr  (ev— ary=  fw,  d  ar  (e—  ar^=4rwr  el (85) 

«-/  W 


, 

Concentrated  loads— 


-         r      **  —     r  ,       v  .., 

A   ]         —r-  -r-  -  P,  (ec—ar)-         ...  (r/) 

''  V.  <•'>•  Vf  ) 


GENERAL  RELATIONS.  11 

Partial  uniform  load— 

OT 

2  Pf  (er—ary=Cwr  d  ar  (e,,-o^  .    .    .  d,<ev (82) 

a,. 

a  r 
1'  Pr  (cv—ary=  Cw,  d  ar  fa—Orl*   •    •    •  a'£ev    ....  (83) 


Uniform  load  over  all— 
e, 

Z  Pr(rr—avy=Civr  d  ar(er—cirY^--^r-wr  ft    •    •  .    .(84) 

J  4- 


o 

c, 


1'  Pr  (<>—a^=  Cwr  d  af  (ev~^Y=-i  wr  c°  .  (85) 

./  ?> 


For  all  loads 


v=l 


r"=2:  A  ^~  ..........  See  (50)  ' 


12  THE  CONTINUOUS  GIRDER. 

/S,.=:(7r+F;)  #,.+, See  (60)  .....  (?») 

frr  =(lr_i  -f  /Vl/)    0r-k(Jr+#)    #r-,       -       -       .    SeC     (61)    .       .       ...       .    («) 

^,._; See  (62)    .    .  .  (0) 

/,_;  /,.  ) 

Or=6  E  f, See  (51) (e) 

c,=o.         ca=l. 

cm-     —2cm_j-^L rm_s  '/"~~      .    .    .  See  (66) (p) 


d.=  -2  rf._^=±L.  ._rfM  |= 

/Jx-i«+2  P«-,>i+2 


.    .See  (68)  .....  (r) 


By  means  of  the  above  equations  we  can  deduce  the  bend- 
ing moment  over  any  support  of  any  continuous  girder  under 
the  single  assumption  that  the  modulus  of  elasticity,  E,  shall 
be  constant. 

E  AND  I  CONSTANT. 

If  the  moment  of  inertia  as  well  as  the  modulus  of  elas- 
ticity is  constant  our  equations  will  be  much  more  simple. 

By  inspection,  we  see  that  (a)  equals  zero,  and  hence  jPr, 
Fr  and  F,,'.'  equal  zero  and  also  Hr  and  Ht  ;  hence  X,'  and 
X^Lt  equal  zero.  By  reduction  we  obtain 


.      (A,) 


GENERAL  RELATIONS.  13 

r,=o.     rg=--l. 

J    ,?     .          /,„-*+/,„-/  In,-, 

Lltt ;  'ill  — I 

(l—o.     da=l. 

lS-m+3    +     lg-m+3  l 


-m+3  fc.<— m+S 


Concentrated  loads— 

A  =    -  I  Pr  II  (2  kr—3  #  +  fcj) (v;;) 

Partial  uniform  load  — 

Ar=    -  wr  /;  ]  ^-4?  h-j-  r (86) 

4  J  «;:  ==ibr  /,. 

Uniform  load  over  all— 

Ar=  ~  wt.  /; (87) 

"^  » 

Concen  t  rated  lott  ds — 
Partial  uniform  load— 

{O     L2 7./,  -j    (tr=Kr   lr 
•%r£  (88) 

j  o;=kr  ir 


Uniform  load  over 

4rW* (89) 


14  THE  CONTINUOUS  GIRDER. 

E  AND  I  CONSTANT— Span*  equal. 

If  the  spans   arc   equal,  the  equations   of  the   last   case 
reduce  to, 


«-•«+*  *  '    1  H  ) 

r/,-0.       <k=l 

dm=    -  4  d.,,-*  —  (I*-, (rs) 

Ar 

Concentrated  loads— 

Ar=    -  I  Prl*  (2  kr—3  A^-j-Ar;) ;.<«,) 

Partial  uniform  load— 

7U     -j    «r-     /,'..   / 

/*  J    z^      z.»  i        ''    I  /qn\ 

~7~ (yu) 


Uniform  load  orer  all—  . 

^r=  -«v> (91) 

B, 

Concentrated  loads— 

B,.=   -  1'  P,.  I3  (k—ty (jg) 

Partial  uniform  load— 
t'etr'  &.*  o'r=krt 

*=  -^'l^Tr).  •••(92) 


GENERAL  RELATIONS.  15 

Uniform  load  over  all— 

Br=  -  wr  F .  (93) 


-y 


—  h 

V  I'    T?  f  ^  "     "r       "Jr — '       "(•+/ 


In  case  the  supports  are  at  the  same  level,  Yr=o  in  all 
equations. 

From  (p.})  and  (/*.,)  we  obtain  the  following  values  for  c 
and  (I: 

Ci=  ±       0  =  d,  c7  =  780  =  d, 


c3=  -       4  =  ds  c9  =          10864  =  d0 

c4=  +    15  =  d,  C«F=  +  .  40545  =  dlo 

c5=  -      56  =  d,  Cll=    -  151316  =  dtl 

c6=  +  209  =  clG  r,3=  -f  564719  —  d» 

Explanation  of   Table  I  —  In   the   co-efficients 
(2  kr  —  3  #  +  #)  and  (kr  —  k?),  k,  is  a  fraction  and  equals 

-y1,  henc^,  it  is  immaterial  about  the  actual  values  of  a 

';• 

and  I  as  long  as  the  ratio  is  known. 

The  ratio  kr  has  been  assumed  to  have  all  possible  values 
from  .001  to  .000  inclusive,  and  the  respective  values  of 
the  above  co  efficients,  carefully  computed,  are  given  in 
Table  I.  A  few  trials  will  prove  the  great  utility  of  this 
table,  and  convince  the  computer  that  much  time  and  labor 
can  be  saved  by  its  use. 

SHEAR. 

The  moments  over  the  supports  being  determined,  the 
shears  can  be  found  from  the  folio  wing  formulas,  which  apply 
to  all  cases. 


16  THE  CONTINUOUS  GIRDER, 


+  <2U     .......  See  (11)  .    .    .    .  (C) 

,._; 

In  which  Q  and  Q'  have  the  following  values: 
Concet  it  MI  ted  loads— 

Qr=?  pr  (i-k,)  ..................  0)4) 

Qf=SPrk  .....................  0)5) 

Partial  uniform  load— 


'  /        / 

Qr=Wr    I,   ]^=^j-         _    '    "  .     .     . 


(  vi  •,   d,=k,  I, 
,.=w,  lr  \-f       „  .    .  (97) 

V  '         Al       /'•        / 


a"r=kr  I,. 
Uniform  load  over  all— 


INTERMEDIATE    BENDING    MOMENT  8. 

General  equations  for  all  possible  cases. 

M^Mr+SrTr—Lr     ..........    See   (8  )    .  .       -    (D) 

In  which  L  has  the  following  values: 

Concentrated  loads  — 
Lr=IPr(x—ar)   ..........  a£x,  ......  (100) 

Partial  uniform  loads— 


Lr=wr     ar(xr-^-}  .....    a'r<z,  ......    (101) 

'  ar=a"r 

(                 n        )  flr—a'r—xr         1 

Lr=wf     ar(yr—-£)  \  .....  -     r  n-r,-l  .....  (102) 

'  a'  ==<£=> 


GENERAL. RELATIONS.  17 

DEFLECTION. 

General  formula.     E  alone  constant. 

?  =h+t      +  :-*—$SM  T*     s.  tf-v  P(x—  aY\ 
6EIt*  » 

+  —  -  "^  (-**-   -  A)    \3Mre   *fc    ,-  H^V  ^    « 

tf  E/I  «,•/„_,  /„       *'  ' 

-rpr  (ey— a>y— 3  (x—ev)  2Pr  (e—ary\    .  See  (39)  .    .  (E) 

tr+l=.  Ad_p^!L  _}_  .__!_  |  Mr  ^_|_  ^  j/r+/  l^^p^l  (kr—ls*r)    | 

1  r  f 

rf        —7—    -'  1-^ -T-||     Mr£  i$—   -T1)-}-®  Mr+1^~ 

6EIJ,  v=lr  ^jv_t       j   '  (  +  * 


+  -f  ?  Pr  lr  (l-kr)^  Pr  (ev-afy-3ev  S  Pr  (e-a,^     .  .(45) 

For  uniform,  loads— 

ar=ar          a,.=kr  lr 
Z  pr=  CWr  d  ar=  Cwr  lr  d  k,  ...........  (12) 

ar=a'r          a'r=kr  lr 

Remember  that  in  the  terms  containing  (x  —  a)  and  (e  —  a), 
x  and  e  must  never  be  less  than  a,  or  rather  a^x  or  e 

E  AND  I  CONSTANT. 

If  the  moment  of  inertia  is  constant  the  preceding  equa- 
tions reduce  to, 


yr=hr+trx,+  ~jjnTrTr-\Srxt~^Pr(xr-~a1):i      .    .   (E,) 

tr+^  —  'f—  *  ^J  {    Mr  1+2  MT+,  lf+S  Pr  l>r  (kr~K)    }   •    -     -   (45a) 


18  THE  CONTINUOUS  GIRDER. 

For  uniform  loads — 

ar=ar          ar=k,  I,. 

Z  Pr=  Cwr  d  Ur=  Cwr  lr  d  kr (12) 

ar=a'r          a'r=kr  lr 

If  the  supports  are  at  the  same  level,  the  terms  containing 
h  in  the  above  equations  become  zero;  the  equations  re- 
main the  same  in  every  other  particular. 

We  have  now  deduced  general  equations  by  which  we  can 
determine  the  bending  moments,  shears  and  deflections  for 
any  continuous  girder.  In  the  next  chapters  we  will  give 
numerous  examples  or  special  cases  illustrating  the  applica- 
tion of  the  formulas. 


II. 

SUPPORTED  GIRDERS. 

CASE  I. 
A  simple  girder  resting  upon  ttvo  supports— 


'Jr.  sfe 

ft  &     W 

£%.  s.       y 


(  a  )     E  alone  constant. 
From  (A]  we  have  for  M,  and  MI2 
t=M3=0  ....................  (103) 

From  (B)  and  (C) 
,-=Qt  ......    .    ..............    .   (104) 

=e;  ......................  (105) 

Or 

,=£  P,  (1—  kt)    .    .   for  concentrated  loads  ......  (106) 

i=2  PI  k,  .    .  for  concentrated  loads  .........  (107) 


Sf=Wt  I,    -,&/—  .    .    .for  any  uniform  load  .    .    .  (108) 

*  !   a'^k,  I, 

(  k2,  )    ai~kt  lt 

S2=wi  I,    \-jrr    •    •    •    -for  any  uniform  load  ....  (109) 
'   d;=k,lt 

$,=  —  wt  lt  ,    .    .for  uniform  load  overall  .    .    .    .    .    .  (110) 


20  THE  CONTINUOUS  GIRDER. 

Sg=  -r-  w,  I,  .    .    .  for  uniform  load  over  all  .    .    .    .    .    .(Ill) 

From  (  J>)  the  bending  moment  at  any  section  x,  is 
Jt'<=$  &,—  L,     ..................  (112) 

Substituting  the  values  of  &,  and  L,  we  have, 

For  concentrated  loads  — 
MX=Z  P<  (l-kt)  x-Z  P,  (x-a,)  .    .    ..  :/<.c.    .        .    .(113) 

For  any  uniform  load  — 

a',  a,=a'i 

M*=  *<  w,  I,  {  k~  f-|      -  w,    j«,  (z,  -  ^  )   j-  a,<z,       (114) 
a/  a,—  a',' 

For  uniform  load  over  all-  — 

1       1  1  1 

Mx=-g-Wi   I,  x—  —wtx2,=  —  w,  xt  (l<—  x,)     ....    (115) 

For  a  uniform  load  over  all,  the  moment  at  the  center  of 
the  girder  becomes, 


The  well  known  formula  for  this  case.  We  see,  therefore, 
that  our  formulas  are  perfectly  exact  for  the  simple  girder, 
and  also  that  a  variable  moment  of  inertia  or  difference  of 
level  of  the  supports  does  not  effect  the  values  of  the  bend- 
ing moments  or  the  shears. 

From  (  JB7),  we  have, 


6  E  7X 

v=l 

—  ? 


-r  P,  (^-a,r-^  (fv-^)  ^P,  (c  -a,)  ~       (117) 


SUPPORTED  BEAMS. 


h  —  h, 


21 
.    (118) 


(b)     12  and  I  constant. 
The  moment  and  shear  equations  are  the  same  as  in  (a). 

y,=h,+  tlr,  +  -six'l-SPl  (z,-o,)*         .    .    .    .  (119) 


IF  the  origin  is  taken  at  one  of  the  supports,  the  corre- 
sponding value  of  faj  will,  of  course,  equal  zero. 

CASE  II. 
A  beam  continuous  over  three  supports  — 


' 


(a)     IE  alone,  constant. 


From  (yl 
M,=M3-=0 


^=o.       c2=L        c3=- 

-  P2 


(120) 


(121) 


(122) 


,=o.      d,=l.      4=    -  ...........    (123) 

PI 

Substituting  (122)  and  (123)  in  (/«/),  it  reduces  to, 


"Pi 


(124) 


22  THE  CONTINUOUS  GIRDEU. 

x,;=n,  0,—F,;  s  p,  i,  (i—kj  o,  ...........  (125) 

x;=  -H;o,-2F;'  zpti,  (/-/,-,  K 


2 


^(<^:^!  />  (,,-^.    ..(127) 


v=l 
;=  Z      A  I  ^  Pt  (e°~a')3  -  3-£  v  p,  (e-a,Y  \     .    .   (  128) 

V     {  I,  i,  ) 

v=l, 


-T-  ..........  •  ........    (129) 


A=v-     -T- (180) 


...........    (183) 

v=!2 


=6  E  /(/,+  F;')-\-6EIt 

.    (134) 


SUPPORTED  BEAMS.  23 

A%  and  BI 
Concentrated  loads — 

A,=    -  ?  P2  II  (2  k,—3  kl-\-kt)  6  E  It (135) 

Bt=    -  S  P,  If  (k—kf)  6EIt '   (136) 

Partial  uniform  loads— 


.  d,=k,  I, 

/?,=    -wJ^-^^^EI,    .    .    .  .    (138) 

a'i  =k,  I, 

Uniform  load  over  all— 
A2=        -J£,w,li#EJt (13(,)) 

n,=  -  -4- Wl  i?  6  E  it  .  ....  (iio) 

4 

a,.=a'r          a'r=kr  lr 
Z  Pr= Cwr  d  a,=  Cwr  lr  d  k, (12) 

"  " 7          7 

Shears  and  intermediate  bending  moments— 

For  shears  and  the  intermediate  bending  moments,  apply 
the  general  equations  (IB)  (c)  and  (I>),  and  for  deflection 
use  equations  (E)  and  (45). 

(6)     JE  and  I  constant. 

In  this  case,  we  have  from  the  equations  under  (a), 
M,=My=o  .  (141) 


24  THE  CONTINUOUS  GIRDER. 

Y,+A,+B, 


: 


In  which, 


Concentrated  loads — 

=    -  i' P, // (Jg  fc-5  */+Jk/) (14*) 

,=    -  S  P,  If  (k,—k?)    ...  (145) 

Uniform  load  over  all— 

<*'a=ks  L 
(146) 


1  *;=*,  /, 

Uniform  load  over  all— 

A2=        4-^  V  ........  (148) 


Bt=  -  W7/  ^  ...................  (149) 

4- 

As  this  is  a  case  much  used  in  practice,  we  will  give  ex- 
pressions for  the  bending  moments  in  terms  of  the  spans 
and  the  loads. 


Concentrated  loads— 

v  _  VP  /  2  K  _  v  P  1  ~  1C 

_  J  *     -  '  r*  '2    *^2     -1  r>  l'i    A/ 


^r\\ 


*(/,+« 

In  which, 
2=2  k,—3  k/+k,°     ....',....,   ......  (151) 

And 

;=k,—k;s    ................    .....  (152) 


SUPPORTED  BEAMS.  25 


From(J?)  and  (  C), 

y  ___  \'  p   l  2   L^  _  v  p   /    £•" 

[+- 


_  -,,, 

~in(-  '  "  •          •  (164) 


Or, 


Uniform  loads  over  each  span— 


Yg—     —      WV       //—      —     W7;       If 


-  F2  +  —  w»2  Z/-(-  —  w,  If 


Or, 


The  above  equations  at  once  reduce  to 

(158) 


—  —w,  II—  —  wf  If 


F2+  -y-  w,  4*+  -y  W,  f 

'  +     -  w<  ''  .....   (160) 


26  TIIK  CONTINUOUS  (J 


,-  -    w,i*-  -w.  i 


(c)     E,  I  and  It 

In  this  case,  the  formulas  are  the  same  as  in  case  (ft)  with 
the  term  Yg=0. 

(d)     E,  J,  h  and  I  constant. 
Making  l,~lg,  the  equations  of  case  (6)  reduce  to 
Conceit  t  rut  i*<1  font  Is 

v  /'   /    l\        v  /'  /    A" 

-  -    /y   h  Ay—-    1  t  i,  A, 


Or, 

V  />       /.'  N'   />       /.' 

- 


.' 

V 


—  , 

v  /'    A'       v  P  V 

3,=  ^-          '-+£P,(1    k.)=R,.  .(166) 


(166) 

/,• 

i    >'  />    A'       v  />  #" 

,  SP.(1    /.,)  '  (167) 


v  /'    /•        v  /'   (  I      1-  \       l>  i  ir 

_/,/,,        _    /      (  /       /,  .  i        /,         .      .    (  I  (>, 


I  it  i  form  loffd  owr  it/l—tr,     //      //• 

'.=  —£-"<• <170> 

,  /  1.5 

-f    —  w;  ^  -^- 


/     i                   /                          /  ^  1 7'^  \ 

l      ~U~  «'  fc    ~T    ~^~  W  I    -        -Q-  V     t  ^  I  /  • )  ^1 

Q  A?  O 


H 


x  f     _         p 

>      W    f  »r  f     -  fV 


n 

t!74) 


.-I 


CASE  III 

or<>r  four 


»,   «J 


( (1 )       /£ 


>      :       v       V        \    •    •       B,rf.) 

.1     )      \      \   i  „•     «  .;  \ 

',.1       )•      .V       \  ../?.<,}    . 

T*  I**    (  ' 

Which  minors  to  tho  following: 
f  ( A-    }     \     .V 


,:  .    -      (        ;       ,        ,        ,  , 


-        - 


f     - 


flTS) 


(176 


"        X 

7>>         I 

:        •  '  \   \ 


I'so  general  e<|iuitions  tor  tho  various  torms  in  the 
and  also  tor  shear,  intormoiliato  moments  anil 


28  THE  CONTINUOUS  GIRDER. 

(b)     E  and  I  constant. 
Our  equations  now  reduce  to 


M       (Y,+A,+B,)  2  (k+g-^, 
4  (ft  ' 


,.       ,-, 

4  (/,+«  (4+4)-4* 

(c)     H,  I  and  h  constant. 

2  (4+4)  (^+B,)-^  (AS+B,) 
- 


TUT    v  \"/^   *g/    V-"an    ^~V       frg    ^x^-p^y  /ioo\ 

«*s  —  — /   //    \i\fi    i   7~\ 71 —  (^"^/ 

(d)     JEJ,  jT,  /i  and  £  constant. 


(184) 


Uniform  loads— 

If  each  span  is  covered  with  a  uniform  load,  we  have 

=  •-*  "'^'-  *-•+*•  tf  .  .  .  .......  (185) 

^-^tf-tf+u,,//     .....    .....   (186) 


If  w,=Wg=ws=w,  and  //=L=/V=£)  we  obtain,  at  once,  the 
well  known  forms, 

M3=  *?  ......    •    ...........  (187) 


-wP  ..................  (188) 


SUPPORTED  BEAMS.  29 

And  for  the  shears  we  have 
=     -^-"M  -  ±-wl=;^.wi=Rf    .  ,    .(189) 

wl  .........  (190) 

Jowl 


(192) 


wi  +  --wi  =  j-wi  ......  (194) 


-ivi-\--wi  =  -v>i*=R4.  ......  •  (196) 


CASE  IV. 


*  We  will  now  consider  the  case  of  the  "Tipper,"  or  a 
beam  continuous  over  four  supports,  the  second  and  third 
supports,  resting  on  a  rigid  beam,  supported,  generally,  by  a 
single  support  at  its  center. 

It  is  evident  that  if  supports  2  and  3  are  supported  by  an 
unyielding  bar,  which  is  supported,  in  turn,  at  its  center,  that 
the  reaction  J?.,  must  equal  the  reaction  It,3,  and  also  that 
-\-fis=  —  Tis.  If  the  unyielding  bar  is  not  supported  at  its 
center,  H,,  and  _B.V  will  be  inversely  proportional  to  their 
lever  arms  around  the  point  of  support,  and  -\-ha  and  — h3 
will  be  directly  proportional  to  the  lever  arms. 


See  "  Strains  in  Framed  Structures,"  page  2o4,  by  Prof.  DuBois. 


30  THE  CONTINUOUS  GIRDER. 

The  unyielding  bar  supported  at  its  center — 


a_^     i 

M,  M,      ^Tf/z>  I        ^ 

'*™^  x>fts*      v?  < 

HP  *  ^ 

^r  •***  * 

~2  I  ~  \**z 

7> 

(a)     E  alone,  constant. 
From  our  general  equations,  by  reduction,  we  obtain, 

''/  ^8 

p  -     ~    3  -L      *       3    i  r>'  -i_  D  n  QK^ 

i  i  V^9  "|     ^C3 V  •*•  vOJ 

Since  R3~R3)  we  have 

— ^ -?-  -f  -    —^1  — i-  +  Q'iJrQ2—Q'2—Q3==0       .     .   (199) 

Q Q' —-V p^  /j ^  fc0\  (200) 

Let 

Q't+Qa—Q's—Qs=Q    -   •   • (201) 

l,=n  I,  and  ls=m  I,    .    . (202) 

Then  (199)  becomes 
n  M3 — m  Ms-\-2  m  n  (M:i—M2)-}-m  n  Q  12=0     .....  (203) 

Or, 
n  M3—m  M2-\-%  m  n  (M3—Ma)=    -  m  n  Q  L (204) 

By  inspecting  the  equations  under  Case  III,  we  see  that 
the  moment  equations  may  be  placed  under  the  following 
forms : 

MS=C;  K+C;  Y,- 


SUPPORTED  BEAMS.  31 

From  (205)  and  (206),  we  have 


;—ct)  ....  (20?) 

Substituting  (205),  (206)  and  (207)  in  (204),  it  becomes 
f  +n  C'2  Y2+n  C'3  Ya+n  C't  } 

I  _TO  a  Ff-ro  C,  Y3-m  C,  [=-mnQl2.  (208) 

+<mn'<<5-G)  F2+£mn(0;-Cv)  Fs 

-f^mn(C;-COJ 

Or,  by  placing   F  and   F'  in  place  of  the  co-efficients  of 
Yj  etc.,  we  have 
VY2+V'Y3=  -mnQl—V"  ........    .    .    .(209) 

(210) 


Hence, 


But,  h2=  —h;f,  or,  h3=  —h»;  therefore, 

v  Y*=  ~  "'  °V    ~    ~v    "'  "*h*  '  '  (212) 


o,k,..  (214) 

Or, 
V  Y3+VYf=U]0a—[2-]  oaha    .........    .    .  (215) 

Therefore, 
[l-\02-[2]02h2=  -mn  Ql—V"  ..........  (216) 

And 


+m  nQ  l,+  V"a 

ffl   02  ~  ............    (217 


32  THE  CONTINUOUS  GIRDER. 

From  (217),  we  can  find  the  value  of  //.,—.  —  //.„  and  then 
the  values  of  Y2  and  Y3  from  (210)  and  (211),  whence  we 
can  determine  Jfi  and  j&f,.  from  (177)  and  (178). 

The  easiest  method  of  finding  the  values  of  the  terms  in 
the  equation  giving  the  value  of  hai  is  to  substitute,  for  any 
particular  case,  the  values  of  all  the  known  quantities  in  the 
immediately  preceding  equations,  or  find  the  values  of  the 
constants  C1}  CX,  etc.,  and  in  turn,  substitute  them  in  the 
equations  containing  them.  The  operations  are  long  and 
exceedingly  tedious,  but  simple,  as  an  examination  of  the 
equations  shows. 

(b)     E  and  I  constant. 

If  the  moment  of  inertia  is  constant,  the  deduction  of  the 
constants  C,,  C2,  etc.,  becomes  much  more  simple.  After 
Jt>  is  determined  from  (217),  and  Y2  and  Y3  from  (210)  and 
(211),  the  bending  moments  are  readily  obtained  from  (179) 
and  (180). 

If  hi=h^=oi  as  is  usually  the  case,  the  process  becomes 
still  more  simple. 

(c)      E  and  I  constant.      h,=h4=o.     1,-l^L     L=n  I.     h»=  -h:i. 
We  have,  as  in  («),  by  a  little  reduction, 

n  (M3—M,)-\-2  (MS—M3)=   -nlQ    .    .    .    .    .....  (218) 

Or, 

(n+2)(Ms-M,)=   -nlQ.    ........  .    .(219) 

Letting  .£  (/,H-4)  <^+4)—  l,*=D  I    .........  (220) 

We  have,  from  (179)  and  (180), 


*,  _t  tt,,   n 

—    — 


s 

Then, 
D  (Ms~M,)=(Ya+As+Bs)  2  (n-^l)-(A^Bt+Ys)  n 

(Y3+A3+B,)n          -(At+Bi+Y,)  2  (n+1)    .    .   (223) 


SUPPORTED  BP;AMS.  ;>-> 

Hence, 

Ya(Sn-\  2)     Ya(Sn    2}  •  (A,  ;  1L 

-AS-B)  (3  n+2)=(M3-M,)  D  .    .  (224) 

Substituting  (224)  in  (219),  it  reduces  to 
(Ys—Ya+Aa+B^A^BdtSn^^fy+fy^  -DnlQ   .  (225) 

Therefore, 

Y,-Y,=   (3~^ln%   -A:,-B3+As+Bl    .    .    .    .(226) 
Since  hj—h4=o  and  hs=  —  h:ij 


n=        6  E  I    =      +  :f    ,  :  +  6  £  /  (228) 

(          I  ('•„>)  (.  M    L  ) 

Therefore, 
1V=    -r,    ....................    (229) 

And 
Y,—  Y2=  -2  K.    ......    .........  (230) 

Hence, 

r  J^/(l_      .  ^.-F-ft-^-^i 

- 


But,  D  1=4  (/,+«  (^+y-//      /;  44-4  4  I, 

^^  I  lt+£    ll_ll=^  /*._)_£  n  ^_|_C5  ^  ^-'^p  (3U3 

l\($n+2)fy+2)  .....  (232j 


And 
Q=^^~({<~Q,    ..............        (201) 

Substituting  (201)  and  (232)  in  (231),  we  have 
Y  _  llM+Qi-Q-Q^A^Bi-At-B,  „ 

•*  2  —  ,>  *3     •       •      I  -'•Jd  ) 

(233)  completely  determines    1^  and    3^y,  and  now  the 
bending  moments  can  be  deduced  from  (179)  and  (180). 


III. 

BEAMS  WITH   FIXED  ENDS. 

In  this  chapter  we  shall  consider  beams  with  one  or  both 
ends  fixed. 

CASE  I. 

A  beam  flxed  at  one  entl  <m<l  stijtjtortcd  at  the 
other — 


For  convenience,  the  left  end  will  be  considered  as  fixed. 

This  case  is,  in  reality,  the  same  as  Case  II,  in  Chapter  II, 

/ 
considering  l,=o,  h,=h.>  and  7i— o,  hence,  we  can  use  the 

equations  of  that  case  by  making  the  proper  changes. 

(a)     E  alone  constant. 
(124)  becomes 

M>-(A+K4A7) (234) 


BEAMS  WITH  FIXED  ENDS.  35 

(134)  becomes 
*  K=       3  n  FJIt  7,    j^r^j    ...........  (235) 

By  inspecting  equations  (124)  to  (134),  inclusive,  it  is  seen 

i 

that  we  can  cancel  out  6*  JEJ  It  Irom  all  the  terms  in  (234), 
hence,  we  can  write 

*F,=   -0j£/,^r^    ..............    (236) 

fi=ls+Fs  ....................    (237) 

2  2 


I,-,         I 


*i  3 

L 


<129> 


v=l 

=^  (133) 


..(127) 

f  ) 

v=L 

;=H3—F:Pals(l-ks)  ..............  (238) 

Concentrated  loads— 

,=  —  I  P,  I*  (2  k»—3  kf+kf)  ...........  (140 


*  In  reality,  7{  ]  ^-,—       —Ii  \  -jr  \  which  is  indeterminate, 

(.  h  }  (.      U~'      ) 

but  the  above  form  is  the  only  logical  one  which  the  ex- 
pression for  Ys  can  take. 


36  THE  CONTINUOUS  GIRDKH. 

Partial  nit  i  form  loads— 

f  T,  A     \    GLo^Ko   I.) 

A3=    -  w,  //  -*/--*/+  --f-         ...........  (146) 

-4  >«:-L/, 

Uniform  load  over  nil— 

A.=        4-  m,  I*  .  .  (148) 

4 

Shears  and  intermediate  betid  itt-f/  moments— 

For  shears  and  the  intermediate  bending  moments,  apply 
the  general  equations  (.B),  (  C)  and  (_D),  and  for  deflection, 
use  equations  (JE7)  and  (4o). 

(6)     JS7  and  JT  constant. 

For  this  case,  (129),  (131),  (133),  (127)  and  (238)  be- 
come zero,  and  (237)  reduces  to 

&=J,   .....................   (239) 

we  have  from  (234)  or  (142) 


In  which  the  values  of  Y2  and  A,,  are  given  by  (236),  (144), 
(146)  or  (148). 

As  this  is  a  case  quite  likely  to  occur  in  practice,  we  will 
give  expressions  for  the  bending  moments  and  shears  in 
terms  of  the  loads  and  span. 

Concentrated  loads  — 

Substituting  in  (  144)  in  (240),  or  from  (150),  we  have 

^=r'~^//g'  ............    ....    (241) 

In  which  K2=2  k,—3  H+H. 
From  (B)  and  (  C),  or  (155), 

S,=  r  f*  ^  K*  4-  I  p^  (1—  ]<„)=&  ......       (242) 


BEAMS  WITH  FIXED  ENDS.  37 

Also,  (see  157). 
fifr=?        -^P^  +  -  A  **=#, .  (243) 

Uniform  load  over  all— 

The  above  equations  at  once  reduce  to 

K-  4-  v*K. 

;!/„=-" 


!>  I, 


y,    ^w,  u 

s?=        ~^rp—       -  +  -J-  ™*  ^=ft 
+;T*  .-    4-  tr.  I* 

s:,= __i_         i 


(c)     .fJ,  I  and  h  constant. 

In  this  case  the  formulas  are  the  same  as  in  case  (b)  with 
the  term  Y3=0. 

The  formulas  are  : 

Concentrated  loads  — 

M        -  Z  1\  I,  K., 

M<=        ~J^  ...........   (247) 

V*=^-j^  -f  •  R  (1-L)=R,    ...........  (248) 

S=    ~—r~^  +?RL=R:i     ......  (249) 

Uniform  load  over  all— 

(244)  becomes 

M*=        --  ?'V//  -    ...............    (250) 


(251) 


38  THE  CONTINUOUS  GIRDER. 

S=     -  4"  Ws  k+  4-  ws  4=  -|-  w>.o  4=#,  ......  (-J52) 

From  (18),  we  have 
JM^:-|-'«b(^t>--tf-4aS)    ......  -    ..(253) 

Mix.  Af=  -    4-  »'"  f  *=^»  ....'•• (254) 

x  <J 

From  (12,)  and  (4  o  a),  we  obtain,  if  7*.=^,— 0, 
y,  =       -  482EI  ws  4  (3  tt—5  I  y^+2  xr!)  .    .  x*=a*  .    .   (255) 

Special  case— 

A  single  concentrated  load  at  the  centre  of  the  beam. 
By  applying  Table  I,  (250)  becomes  at  once 
M*=    -P,  1,0.1875=       -jfiP,l* (256) 


«=      -  jfr  P,  +  —  P,  ==  W  P,  =  «- (258) 

l&=-±  TU  +  ^P^-P,  k-^4  *&*r—  t 

Or, 

M,==  ~  P,  (11  x,—3  L)  .    .   .  x,  <-i-  4 (.260) 

The  maximum  moment  occurs  when  a;,=  -j-  L,  or, 

MaX    M*=:^PJi (2G1) 


BEAMS  WITH  FIXED  ENDS. 


From  (J5J,),  if  ha=h3=0, 


-12  U  x.,-^2  II)  .    .  z,>  - 


And 


CASE  II. 
A  beam  fixed  at  both  ends— 


39 


.    .  (262) 
.    (263) 


This  is  the  same  as  Case  II,  Chapter  II,  with  ht=h3,  h4=hs, 

i  3 

lt=o,  1  3=0,  and  7i=o,-and  It=o. 

(a)     ^J  alone  constant. 
From  (177),  we  have 


(264) 


From  (178), 


•    •    (265) 


40  THE  CONTINUOUS  GIRDER. 

(264)  reduces  to 

And  (178)  becomes 


.  (267) 


TY-y      r        , 

4  ft  ft  —  ft  ft    (. 
In  which,  after  dividing  by  0,^=0—  os, 

(268) 

........  (269) 

(270) 

......    (271) 

.....  (272) 


-.     ..............  (278) 

The  values  of  the  remaining  terms  are  easily  found  from 
the  general  equations,  remembering  that  H,  or  0,,  has  been 
cancelled  out  of  each  term. 

(  b  )     JE  and  I  constant. 
From  (266)  or  (179),  we  have 


From  (267)  or  (180),  we  obtain 

(1-+^)^L-(A+K)  /,_^(}>^)-(A+K) 

5//  "5T"          '  '  (2lo) 

In  which  the  terms  have  the  values  given  under  (<f). 


See  note  under  Equation  (235). 


BEAMS  WITH  FIXED  ENDS.  41 

(c)     JE,  I  and  h  constant. 

(274)  becomes 

M—  'L^?JL (276) 

(275)  becomes 

*?>~^ir- (277) 

For  concentrated  loads  we  can  write 

MS  =  -  3  '  ~  • (2"  8) 

And 

—  2  VP  I  K'4-^P  I  K 

if ~  " r*  l°-  A3   i  ('27Q} 

aJL&  —  « \  £  i  u  I 


Uniform  load  over  all— 

M3=     -  ^  w,  U=M:1 (280) 

SM=    4"    ^^V^S C281) 


2 
If  x=0 


-  12    \ 


If  h^=h ,=- 0. 


1 
Max.     Mx=     -  —^  w,  %=M9=M3 (283) 


=?£-g-j  ™*x**  W—2  '^v+i'/)   -    -    .*,=«:   •    •    .    -(284) 

A  single  load  In  the  centre  of  the  beam— 

By  using  Table  I,  we  have  at  once  from  (278)  and  (279), 


~  P*L-  .    (285) 

A 


(286) 


42 


THE  CONTINUOUS  GIRDER. 


Also, 


*-w  •  •  •  **<  - 


.    -  (287) 
.   (288) 


If  jra=0, 


And 
Sfb= 


«r-*.a5r)    -    -  iKtfrF  /,  -    -    (291) 


CASE  III. 

A  beam  fixed  at  one  end,  and  unsupported  <it 

the  other — 


As  the  right  end  of  the  beam  is  unsupported,  there  can 
be  no  reaction  S3=R3,  therefore,  by  (11), 

Concentrated  loads— 

S3=      ~~ — -  +  -  P>  L=0,  and  since  M3=0,  we  have 


BEAMS  WITH  FIXED  ENDS.  43 

M,=    -IP2Lk,   .    .    .Sa=SPa (292) 

Showing  that  the  moment  of  inertia  does  not  enter  into 
the  expression  for  the  bending  moment. 

Any  uniform,  load— 

M,=        ~  w,  (a-al}  (ai+ai) (293) 

S2=w,  (ai— CQ (294) 

From  (8), 

Mx=0.        x£a2 
And 

Mx=  I  P2  (x  —  a2)  .    .    .  xg<ag (295) 

Also,  for  any  uniform  load, 

Mx=w2  (a—a2 )     x—  a*+a*   1   .    .    .   X£a; (296) 


For  deflection,  use  the  general  formulas,  if  the  moment 
of  inertia  is  variable. 

If  the  moment  of  inertia  is  constant,  we  have,  from  (JEJ,), 
If  h,=0, 

1  c  ~\ 

.    .    .  '297) 


Or,  for  a  single  concentrated  load, 

........  (298) 


And  if  the  load  is  at  the  end  of  the  beam,  y-3=a2=l2,  and 
(298)  becomes 

P  I3 
ya=  --      •  '  *        .    .at  end  of  beam  .........    (299) 

Also,  for  a  uniform  load  over  all,  (xs=ag=la)t 

.........  (300) 


44  THE  CONTINUOUS  GIRDER. 

CASE  IV. 

A  beam  on  ttro  sH/tports,  find  one  etui  nnxttp- 
•ported— 


In  this  case,  JLT,,  Jf£  and  8y  equal  zero,  hence,  from  (11), 
or  (C),  we  have 

S=4^  +<M==^,w,JC=    -ft/,   .........  (301) 


Which  is  precisely  the  same  equation  we  obtained  in  Case 
III.     The  values  of  Q',  are  given  in  (95),  (97)  and  (99). 

From  the  general  equations  (B)  and  (  C  ),  we  obtain 
S,=     .  +  QI=R,  .................  (302) 


S=  -y^     +  ft  1      - (303) 

V  H 

5,=  -|P   -fft  J (304) 

From  (D),  we  obtain 
^=8,  x,  —  L,=    -y-^  -f  Q,  x;  — L; (305) 

^^-^  4-  ft  sc»— 4   ....  (306) 


Note  that  the  moment  of  inertia  does  not  appear  in  any 
of  the  above  equations. 

For  deflection,  use  the  general  equation  (JE). 


BEAMS  WITH  FIXED  ENDS.  45 

CASE  V. 

A  beam   on    one  support,  having   one  end 
fixed,  and  the  other  unsupported— 


M,          M 


(a)     E  -alone,  constant. 

In  this  case,  as  before,  M4=0,  and  St=0,  hence, 

y=    -  Q3  /,  .....    ..............  (307) 

From  (63),  we  have 

2ft2+2M;ft^=A^B^Y^X^X:  ........  (308) 

Substituting  (307)  in  (308),  it  becomes,  after  reduction, 

...   (309) 


(6)     JE  and  I  constant. 
(309)  becomes 


>*j (sio) 

Ms=    -  Q:i:i (307) 

(c)     J5J,  I  and  h  constant. 

1    c  ^ 

(311) 


M,=    -.<g  4 (307) 

The  shears,  intermediate  bending  moments  and  deflec- 
tions are  readily  obtained  from  the  general  equations. 


46 


THE  CONTINUOUS  GIRDER. 


CASE  VI. 


A  beam  on  two  supports,  hftriut/  tteitlter 
supported— 


We  have,  at  once,  M,  and  MA  equal  zero,  and  hence,  from 


S,  =   —r^-  -f  Q,=0,  or  Ma=    -  Q,  I, 
And,  from  (C), 


(312) 


(313) 


The  shears,  intermediate  bending  moments  and  deflec- 
tions can  now  be  readily  obtained  from  the  general  equations. 


IV. 
*     THE  POINT  OF  ZERO  MOMENT. 

Let  us  take  (-D). 

In  which  L  is  dependent  upon  the  kind  of  loading  in  the 
span  r.     If  there  is  no  load  in  the  span  r,  then 

Hfx=Mr+Srx, (314) 

Now,  if  there  is  a  point  of  zero  moment  anywhere  in  the 
span  /•,  we  can  find  its  distance  from  the  left  support  by 
making  (314)  equal  zero,  and  solving  for  xr',  doing  this,  we 
obtain 

(315) 


Sr 

From  (JB), 

Sr  =     Mf+l~Mr    4-  ( qr  in  this  case  =  0) (B) 

Now, 

Mm=cm  '?:  /?:!  '    '    '  '3T'    Ms  .    ,    .    .  wO-f.7 (70) 


-\-l  indicating  that  only  those  loads  upon  the  right  of 
the  span  are  considered. 

Substituting  (B)  and  (70)  in  (315),  it  reduces  to 


Cr  —  Cr+t  ~a 

Pt 


(316) 


See  "Annalesdes  Fonts  etChaussees,"  188(5,  PaperNo.  40,  by  M.  Collignon. 


48  THE  CONTINUOUS  GIRDER. 

We  also  have,  for  loads  on  the  left, 


.  And,  substituting  (71)  and  (J5)  in  (315),  we  obtain 

Xr=  -  ^=^  -  ^~     I  .....    LOA,,  ,*  THK   I.KKT      .....       (317) 


s— r+/     ~5 

Pr+f 


(316)  and  (317)  are  general  equations.  Knowing  the 
points  of  zero  moment,  we  can  tell  .at  once  what  loads  must 
be  considered  to  have  a  certain  effect. 

The  values  of  x,.  are  very  easily  computed,  but  they  can 
be  constructed  graphically  as  soon  as  cn  f,+/,  etc.,  are  known. 
Thus:— 


-jx,i  v   *x'H 


>-/'" 

If 


Let  IB  C  represent  any  unloaded  span.     Then,  IB  H=xr 
for  m<,r+l  or  a  load  on  the  right,  and  T>  K=.r,  for 

m>r  or  a  load  on  the  left,  if  E  B=cn  C  F  '=--     -  cr+i  -^ 

Pr 

F  C=d^r+t^  and  B  E'=  -  </,_,+,. 

/";•+/ 

For,  from  Fig.  9, 


I^*H!KH  -V- 


THE  POINT  OF  ZERO  MOMENT. 


49 


Also, 


R  E'  +  CF 


lr=xf  .    .  m>r 


H-/ 


If  the  span  considered  is  loaded,  then  the  value  of  xr  must 
be  obtained  from  (  T)). 

If  the  supports  are  level,  and  the  span  considered  is  uni- 
formly loaded,  .Tr,and  also  the  bending  moments,  can  be 
found  graphically. 


Considering  the  span  r  as  discontinuous  and  uniformly 

1  1 

loaded,  the  bending  moment  at  the  centre  equals  -  -  wr  l'r. 

o 

Now,  in  Fig.  10,  we  wish  to  draw  the  moment  parabola,  so 
that  A  M=Mr  and  B  N=Mr+l.      We  know  H  K  must 

equal  -~-  wr  I2,..     Let  D  and  C  be  the  points  of  zero  moment 
o 

for  loads  on  the  right  and  left  of  the  span  r.      On  the  ver- 
tical H  K  take  any  point  Jt  and  draw  h  D  in  and  7t  C  n, 

and  connect  in  and  n.     Make  n p  =  -  -  wr  ?;'  and  draw 

o 

p  r  parallel  to  in  n  until  it  is  intersected  by  in  h,  pro- 
duced in  the  point  r.    Draw  r  H  and  s  jKparallel  to  li,  and 

also  jM~ ^parallel  to  in  n.  Then  is  H  K=  -~  wr  I2n  A  H>f= 

o 

Mn  B  y=Mr+l  and  A  E  and  A  F=xr. 


50 


THE  CONTINUOUS  GIRDER. 
The  co-efficient*  c  (tit<?  d. 


3:     *,     A 


^*l. 


iuA  — -tT  C       ^-|. ft 


z.H 
•t 


£*# 

The  co-efficients  r?  and  r?  can  be  found  graphically,  if  1 
is  considered  as  constant. 

In  Fig.  11,  let  b  m  and  in  k  represent  the  first  and  sec- 
ond spans  of  a  continuous  girder.  Bisect  b  in  at  A,  m  /»• 
at  B,  and  b  k  at  C.  Make  C  I  -2  r,—  -  2  and  draw 
a  A  ?  e  f,  &  right  line  passing  through  ^1  and  £,  then  will 
J5  e  equal,  numerically,  r...  Thus:— 

but  b  c  h  k=f  A  k — a  b  A=f  n  m  k, 

since  den  =f  e  g,f  n  m  k=g  d  m  k=L  (B  e), 

Therefore,  B  e  —    -2    ',  '*    -.=  r:i. 


tl;i  is  found  in  a  similar  manner  from  Fig.  12. 


k  B=  ^,bA=   lr,kO= 


BEAMS  WITH  FIXED  ENDS. 


51 


b  c  h  k=  -2  d3  (X-fC,)=    -  2  (/S-K_,), 
k  g  d  m=(B  e)  /s_,,  but  b  c  h  k=k  g  d  m, 
Therefore, 


Then,  in  general,  from  Fig.  13,  we  have,  if 


^L if 


p  b  =  c>ll_3  and  1C-    —2  cm_h 
c  b  k  h=d  g  r  d+p  r  k  b=d  g  k  m+p  d'  m  6 
but,  g  d  m  k=(B  e)  lm_,,  p  d  m  b=cm_.>  lm_s 
and  cbkh=    -2  cm_t  (lm_,^lm_2) . 
Therefore, 

L-/  -4-  lm-  lm- 

B  e  =  —  2  cm_, r—  -  cm.a  T—  -  ==  cm. 


And  for  dm  we  can  write 

r>  rt     J  "8 — «-f2         "» — n 

n  e=  --  %  dm_,  •. -. 


L-, 

f's—m+3 


s — w-j-2 


V. 


APPLICATIONS. 

We  will  now  give  a  complete  analysis  of  a  continuous 
girder,  to  illustrate,  more  particularly,  the  use  of  Table  I. 

Ex.  1. — Let  Fig.  14  represent  a  continuous  girder  of  three 
spans  on  level  supports,  and  having  a  constant  moment  of 
inertia. 


jr     -jf     -ft      if    * 

I j. Z  =  300 

i '  £ 


Let  each  panel  be  30  in  length  and  the  loads  repre- 
sented as  in  the  figure.  Then  8=3,  r=l,  2,  3  and  4,  m=l, 
2,  3  and  4,  ll=24D'=la  and  l,=300'. 

First,  look  up  in  Table  1^  the  values  of  the  co-efficients 
2  k — 3  £2-}-F  and  k — k3l  for  the  different  values  of  — -  =  k. 


fc,-*. 

**,-*«+*! 

k-V 

0.123,046,875 
0.234,375,000 
0.322,265,625 
0.375,000,000 
0.380,859,375 
0.328,125,000 
0.205.078,125 

k? 

2k^3li 
+  g 

ky-kl 

0.100 
0.200 
0.300 
0.400 
0.500 
0.600 
0.700 
0.800 
0.900 

0.171 
0288 
0.357 
0.384 
0.375 
0.336 
0.273 
0.192 
0.099 

0.099 
0.192 
0.273 
0.336 
0.375 
0384 
0.857 
0.288 
0.171 

0.125 
0.250 
0.375 
0.500 
0.625 
0.750 
0.875 

0.205,078,125 
0.328,125,000 
0.380,859,375 
0.375,000,000 
0.322,265,625 
0,234,375,000 
0.123,046,875 

APPLICATIONS. 
From  (A,),  we  have  M,=0=M,. 


53 


Ma= 


A> 


From  (p,)  and  (l*/),  we  find  that,  since  c,=0  and  ca=lj 
ca=  —  3.6  and  CA=  -j-  ^-#£. 

Also, 
d;— #  and  c?2=-/,  ds=  —3.6  and  <i4—  -4- 14.93. 

Finding  the  values  of  Af,  A2,  IB,,  _R>,  etc.,  from  (•/,)  and 
(,//),  and  substituting  them,  with  the  values  of  c  and  d 
above,  in  the  moment  equations,  we  obtain, 

TABLE  «— VALUES  OF  Mo. 


— 

Pi 

(0.123,046,875)  ] 

f--    7.111,198 

P] 

— 

(0.234,375,000) 

-  13545,140 

Pf 

— 

Pf 

(0.322,265,625) 

-  18.624,568 

P! 

(0.375,0 

no  ooo^  l    ft? 

7996  —  .    . 

—  91  fi79  99S 

pt 



Pf 

((X380J859|375) 

) 

-  23  010,853 

j  i 

— 

Pf 

(0.328,125,000) 

-  18.963,196 

PS 

— 

PI 

(0.205,078,125)  J 

1-11.851,998 

PI 



Pi 

(0.171) 

f  +  P^ 

(0.099) 

-  12.958,194 

p.i 

— 

PI 

(0.288) 

+  PI 

(0.192) 

-  21.190,636 

p.^ 

— 

(0.357) 

+  P| 

(0.273) 

-  25.389,634 

\>;i 

1  2 



P| 

(0.384) 

4-  PI 

(0.336) 

-  26.247,494 

]>1 

— 

Pi 

(0.375) 

h  90.3010+25.0836  •{  +  P\ 

(0.375)  }-     =  -j  —  24.456,525 

— 

Pi 

(0.336) 

Ipi 

(0.384) 

-  20.709,033 

l;f 



PI 

(0.273) 

(0.357) 

-  15.697,327 

— 

PI 

(0.192) 

+  PI 

(0.288) 

-  10.113,715 

ii 

(0.099) 

I  +  PI 

(0.171) 

—    4.650,503 

•4- 

P^ 

(0.205,078,125)  ] 

+  3.292,221 

pj 

4- 

p2 

(0.328,125,000)  1 

4-  5.267,554 

PI 

-4- 

PI 

(0.380,859,375)  | 

4-  6.114,126 

PI 

+ 

Pf 

(0.375.000,000)  }•  16.0535-  -       . 

......      •{  -f  6.020,061 

p| 

-f 

Pi 

(0.322,265,625) 

-4-  5.173,491 

PI 

-f 

P^ 

(0.234,3 

75,000) 

4-  3.762,365 

pi; 

. 

Pa7 

(0.123,046,875) 

1  -f-  1.975,333 

PS 

THE  CONTINUOUS  GIRDER. 
TABLE  a— Continued. 

VALUES  OF  M,. 


+  Pi 

(0.123,046,875)1 

f  -t-  1.975,333  P} 

+  Pf 

(0.234,375,000) 

4-  3.762,365  Pf 

-f  Pf 

(0.322,265,625) 

4  5.173,491  P? 

(0.375,000,000)  1-  16.0535=   \  -f-  6.020,061  Pf 

f  PI 

(0.380,859,375) 

4-  6.114,126  Pi 

+  Pf 

(0.328,125,000) 

5.267,554  P? 

-rP{ 

(0.205,078,125)  J 

L+  3.292,221  Pi 

-f  Pi 

(0.171)1 

f  —  P£  (0.099)1 

-    4.650,503  PJ 

4-  P| 

(0.288) 

-  P|  (0.192) 

-  10.113,715  PI 

f  PI 

(0.357) 

-  P|  (0.273) 

-  15.697,327  Pi! 

+  P| 

(0.384) 

-  Pi  (0.336) 

-  20.709,0:53  P4 

+  Pi 

(0.375)  \  25.0836+90.3010  \  —  P52  (0.375)  j           -  24.456,525  P5 

f  P* 

(0.336) 

—  P«  (0.384) 

-  26.247,494  P!) 

•4-  P| 

(0.273) 

-  PI  (0.357) 

-  25.389,634  Pi 

-f  P! 

(0.192) 

-  PI  (0.288) 

-  21.190,6:50  V:, 

-f  Pf 

(0.099)  J 

[—  P»  (0.171)  J        1—  15.958,194  P:; 

_P1 

(0.205,078,125)  1 

f—  11.851,998  PJ 

—  PI 

(0.328,125,000) 

-  18.963,196  P^ 

-PI 

(0.380,859,375) 

22.010,853  P^ 

(0.375,000,000)  l  57.7 

926    -....-{-  21.672,225  P| 

pi 

*-  3 

(0.322,265,625) 

-  18.624,568  PJ 

pe 

*•  3 

(0.234,375,000) 

-  13.545,140  Pj; 

(0.123,046,875)  J 

1-    7.111,198  P,7 

Compare  the  values  of  M2  and  BI;1,  and  notice  that  the 
co-efficients  of  JPJ,  JP,6,  etc.,  of  Mit,  are  the  same  as  the 
co  efficients  of  J0/,  J?/,  etc.,  of  JfC,  as  they  should  be,  owing 
to  the  symmetry  of  the  girder. 


APPLICATIONS.  55 

TABLE  b 

The  next  step  is  the  deduction  of  S  from  (J£)  and  (C). 


FIRST  SPAN. 

SECOND  SPAN. 

THIRD  SPAN. 

s< 

K 

8, 

•& 

ft 

8* 

Pi 

+0  845,870 

-i-0.  154,630 

+0.030,288 

-0.008,230 

-p? 

+0.693.5<i2 

+0.306,438 

+0.057,691 

—0.015,676 

Pf 
PI 
Pi 

+0.547,397 

+0.409.WH 
+0.283,288 

+0.452,603 
+0.590,301 
+0.716,712 

+0.079,327 
+0.092.307 
+0.093,749 

Same  as  82 
with—  sign 

-0.021,556 
-0.025,083 
-0.025,475 

Same  as  S  3 
with+sign 

Pf 

+0.170,98<> 

+0.829,014 

+0.080,769 

-0.021,948 

PI 

+0.075,610 

+0.924,384 

+01,050,480 

-0.013,717 

Pi 

-0.053,992 

+0.927,692 

+0.072,307 

+0.019,377 

?! 

-:  1.088,294 

•+0.836,923 

+0.163,076 

+0.042,140 

pi 

-0.105,790 

+0.732,307 

+0.267,692 

+0.065,405 

PI  ' 

| 

-0.109,364 
-0.101,902 
-0.086,287 

Same  as  Si 
with+sign 

+0.618,461 
+0.800,000 

+0.381,538 

+0.381,538 
+0.500,000 
+0.618,461 

+0.086,287 
+0.101,902 
+0.109,364 

Same  as  S3 
with—  sign 

p| 

-U.065,405 

+0.267,692 

+0.732,307 

+0.105,790 

PI 

-0.042,140 

+0.163,076 

+0.836,923 

+0.088.294 

TJ9 

-0.019,377 

+0.072,307 

+0.927,692 

+0.053,992 

PJ 

+0.013,717 

-0.050,480 

+0.924,384 

+0.075,616 

P§ 

+0.021,948 

-0.086,769 

+0.829,014 

+0.170,986 

8 

Pi 

+0.025,475 
+0.025,083 
+0.021,556 

Same  as  Si 
with—  sign 

-0.093,749 
-0.092,307 
-0.079,327 

Same  as  S2 
with+sign 

+0.716,712 
+0.590,301 
+0.452,603 

+0.283,288 
+0.409,699 
+0.547,397 

PS 

+0.015,676 

-0.057,691 

+0.306,438 

+0.693,562 

p 

+0.008,230 

-0.030,288 

+0.154,630 

+0.845,370 

The  computation  of  the  above  table  is  very  simple  and 
easy.     Thus  :  — 

$,—  -y^—  -f-  PI  (.1  —  &/)»  and  numerically  becomes,  for  loads 
if 

in  the  first  span, 


—  7.111  + 


210 


40 


P<=.  -f-  0.845  P/. 


240 


f         -      P?=  +Q.698P*,  etc,  etc. 


For  loads  in  the  second  and  third  spans,  we  have  merely 
* ,  or  the  moment  over  the  second  support  divided  by 
the  length  of  the  first  span. 


56  THE  CONTINUOUS  GIRDER, 

Sg=     —, —      -f-  P2  (1 — L),  for  loads  in  the  second  span,  and 

Sg=     — — -,-  for  loads  in  the  other  spans. 

'2 

-M 

Sv—         *-  -f  Pa  (1 — ks)y  for  loads  in  the  third  span,  and 

83=         2j  for  loads  in  the  other  spans. 

It  is  necessary  to  compute  only  $„  Sy  and  $,  to  fill  out 
the  table  of  shears,  since  S,-\-Sg==0  or  jP,,  Sa+S3=0  or  _P2, 
etc.,  but  it  is  better  to  compute  S',,  S',  and  $,  as  a  check. 


TABLE  r— VALUES  OF  J 


FIRST  SPAN. 


0 

1 

* 

3 

4 

o 

6 

7 

i 

9 

M< 

i 

Mi 

Mi 

i 

MI 

M; 

1 

PI 

+  25.361 

+  20.722 

+  16.083 

+  11.444 

+    6.805 

+    2.166 

-   2.472 

P7, 

2 

P 

+  20.806 

+  41.613 

+  32.420 

+  23.237 

+  14.044 

+    4.851 

-    4.341 

PS 

8 

Pf 

+  16.421 

+  32.843 

+  49.265 

+  a5.687 

-22.109 

+    8.  -531 

-    5.046 

4 

P| 

+  12.290 

+  24.581 

+  36.872 

+  49.163 

-T  31.554 

+  13.745 

-    3.963 

P4 

5 

P 

+    8.498 

+  16997 

+  25.495 

+  33.994 

+  42.493 

+  20.991 

-    0509 

PT 

6 

Pf 

+    5.129 

+  10.259 

+  15.388 

+  20.518 

+  25,647 

+  30.777 

+    5.907 

p-j 

7 

PI 

+    2.268 

+    4,536 

+    6.805 

+    9.073 

+  11.342 

-r  13.610 

+  15.679 

Pi 

8 

VI 

-    1.619 

-    3.239 

-   4.859 

-    6479 

-    8/9S 

-    9.71* 

-  11.338 

p* 

9 

PS 

-    2.648 

-    5.297 

-   7.916 

-  10.595 

-  13.244 

-  15  892 

-  18.541 

Pg 

10 

-    3.173 

-    6.347 

-  '  9.521 

-  12.694     -  15.868     -  19.042 

-  22.215 

PS 

11 

P| 

-    3.280 

-    6.561 

-    9.842 

-  13.123 

-  16.404 

-  19.685 

-  22.966 

12 

PI  !  -    3.057 

-    6.114 

-    9.171 

-  12  228 

-  15.285 

-  18.342 

-  21.399 

PS 

13 

Pi  :  -   2.588 

-    5.177 

-    7.76-5 

-  10.354 

-  12.943 

-  15.531 

-  18.120 

PA 

14 

PA     -    1.962 

-    3.924 

-    5.  886 

-    7.848  ;  -    9.810 

-  11.772 

—  13.735 

P3 

15 

PI  |  -    1.264 

-    2.528 

-    3.792 

-    5.056 

—    9.321 

-    7.58-5 

-    8.849 

PR 

16 

Pi 

-    0.581 

-    1.162 

-    1.743 

-    2.325 

-    2.906 

-    3.487 

-    4.0  9 

pi 

17 

PI 

+    0.411 

+    0.823 

+    1.234 

+    1.646 

+    2.057 

+    2469 

+   2880 

Pi 

18     P^ 

+    0.658 

+    1.316 

+    1.975 

+    2.633 

+    3.292 

+    3.950 

-r    4.609 

P? 

19  1  P3 

+    0.764 

+    1.528 

+    2.292 

-r    3.057 

+    3.82L 

+    4.585 

+    5.349 

Pf 

20     P| 

+    0.752 

+    1.504 

+    2.257 

+    3.009 

+    3.762 

+    4.514 

+    5.267 

Pf 

21     Pi> 

+    0.646 

+    1.293 

+    1.940 

+    2.586 

+    3.233 

+    3.880 

+    4.527 

22     Ptj 

+    0.470 

+    0.940 

+    1.410 

+    1.881 

+    2.351 

+    2.821 

+    3.291 

P| 

23 

£1 

+    0.246 

+    0.493 

+    0.740 

+    0.987 

+    1.234 

-r    1.481 

+    1.728 

3 

3 

3 

3 

., 

3 

3 

m 

m 

M* 

M£ 

Mi 

m 

AC 

THIRD  SPAX. 


APPLICATIONS. 


57 


TABLE  c— Continued, 


SK(  OND  SPAN. 


/ 

2 

r> 

4 

~) 

6 

7 

8 

9 

10 

11 

m 

Ml 

m 

Mi 

M* 

'MI 

m 

Ml 

m 

1  >•_' 

P? 

ft 

!•! 

-    6.202 
-  11.824 
Hi.  241 
-  18.903 
-  19.198 
-  16.540 
-  10.337 

-   5.293 
-  10  103 
-  13.804 
-  16.133 
-  16.385 
-  14.117 
-    8.823 

-    4.385 
-    8.382 
-  11.485 
-  13.364 
-  13.573 
-  11.693 
-    7.308 

-   3.476 
-    6.662 
-    9.105 
-  10.595 
-  10.760 
-    9.270 
-    5.794 

-   2.567 
-    4.941 
-    6.725 

-    7826 
-    7.948 
-    6.847 
-    4.279 

-    1.659 
-    3.220 
-    4.345 
-    5.05(5 
-    5.136 
-    4.424 
-    2.765 

-    0.750 
-    1.500 
-    1.965 
-    2.287 
-    2.323 
-    2.001 
-    1.251 

+    0.157 
+    0.220 
+    0.413 
+    0.481 
+    0.498 
+    0.421 
+    0.263 

+    1.006 
1     1.041 
+    2.793 
+    3.250 
+    3.301 
+    2.844 
+    1.777 

il 

ii 

P 

i:f 
P! 

+  14.872 

+    3.917 
-    3.420 
-    7.693 
-   9.456 

+  12.703 
+  29.024 
+  18.548 
+  10.860 
+    5.543 

+  10.534 
+  24.132 
+  40.517 
+  29.413 
+  20.543 

+    8.364 
+  19.240 
+  32  487 
H   47DU7 
+  35.543 

+    6.195 
+  14.347 
+  24.456 
+  36.521 
+  50.543 

+    4.026+    1.857 
+    9.155'+    4.563 
+  16.425|+    8.394 
+  25.075  +  13.629 
+  35.543'+  20  543 

-   0.312 
-    0.329 
+    0.361 
+    2.183 
+    5.543 

-    2.481 
-    5.221 
-    7.666 
-    9.263 
-    (1.450 

ii 
n 

M9 

ifc 

m 

m 

j\b 

m 

m 

M; 

m 

SECOND  SPAN. 


MX=S,  w,—  !*,  (Wr-di)  for  loads  in  the  first   span,  and 

/ 
Mx—S,  Xi  for  loads  in  the  other  spans.      Mx  =  83  W3  —  IP3 

(w3 — a3)  for  loads  in  the  third  span,  and  JJ£=&  X3  for  loads 
in  the  other  spans. 

If  there  were  no  equal  spans,  then  Table  c  would  have 
the  number  of  apices  squared  computations,  making -its 
formation  tedious,  if  many  spans  were  considered,  but  the 
great  worth  of  the  table,  when  once  computed,  overbalances 
the  hard  labor  in  its  formation. 

Having  Table  c  before  us,  we  can  tell  at  once  which  of  the 
apices  must  be  loaded,  to  produce  a  certain  result,  in  any  par- 
ticular chord  member. 


58  THE  CONTINUOUS  GIRDKK. 

MAXIMUM  MOMENTS. 

Table  (c)  enables  one  to  find  the  maximum  bending 
moments  for  any  chord  piece,  with  comparatively  little  labor. 

r 

Dead  load— 

For  the  dead,  or  static  loads,  of  the  structure,  the  maximum 
bending  moment  for  any  chord  member  is  found  by  taking  the 
algebraic  sum  of  the  quantities  in  the  proper  column  of  Table  ( c) 
(each  co-efficicient  is,  of  course,  multiplied  by  its  IP). 

For  example,  suppose  the  maximum  bending  moment  at 
the  second  panel  of  the  first  span,  or,  better,  the  second 
panel  point  of  the  first  span,  is  desired :  Multiply  each 
co-efficient  in  column  3  of  Table  (#),  by  its  proper  JP,  and 
take  the  algebraic  sum  of  the  products. 

Live  load— 

For  the  live  or  moving  load,  the  maximum  negative  mom<  nt 
at  any  panel  point  is  found  by  taking  the  sum  of  the  negative 
co-efficients  (multiplied  by  their  proper  IP's)  in  the  proper  column. 
of  Table  (c),  and  vice  versa  for  the  maximum  positive  bending 
moment. 

For  example,  suppose  the  maximum  negative  bending 
moment  at  the  second  panel  point  is  desired  :  Take  the 
sum  of  the  negative  products  in  column  3  of  Table  (c). 
For  positive  moment,  take  the  sum  of  the  positive  products. 

If  the  IP's  are  equal,  the  work  is  somewhat  easier,  as  the 
co-efficients  can  be  at  once  summed,  and  then  multiplied 
by  the  common  value  of  IP. 

The  maximum  bending  moments  over  the  supports  are 
obtained  in  the  same  manner  from  Table  (a). 


APPLICATIONS.  59 

MAXIMUM  SHEAR. 

The  maximum  shear  is  obtained  from  Table  (6). 

.J)ead  load— 

The  maximum  shear  at  any  panel  point  is  found  by: 
First. —  Taking  the  algebraic  sum  of  the  co-efficients  (multiplied 

by  their  JP*s)  of  all  the  loads  outside  of  the  span  in  which  the 

apex  considered  lies. 

Second. — Considering  the  span  in  which  the  apex  lies  alone, 
and  using  for  the  left  reactions  the  co-efficients  as  found  in  the 
column  giving  the  values  of  8. 

For  example,  take  the  second  apex  of  the  first  span.  The 
maximum  shear  is  found :  First,  by  taking  the  algebraic 
sum  of  the  co-efficients  in  column  $,  for  the  second  and 
third  spans.  Second,  by  taking  the  sum  of  the  co-efficients 
opposite  JP/,  ff  to  JP1/,  inclusive,  and  decreasing  it  by  the 
co-efficient  opposite  JP/  in  column  S's. 

Live  load— 

For  positive  or  negative  shear : 

First. —  Take  the  sum  of  the  positive  or  negative  co-efficients 
(multiplied  by  their  JP^s)  for  spans  in  which  the  apex  does  not  lie. 

Second. — Consider  the  span  in  which  the  apex  lies  alone,  and 
use  for  left  reactions  the  positive  or  negative  co-efficients  (multi- 
plied by  their  J?'s)  as  given  in  the  proper  column  of  Table  (b). 

For  example,  find  the  maximum  positive  shear  at  the 
second  apex  of  the  first  span  : 

First. — Take  sum  of  co  efficients  (multiplied  by  their  jP  V) 
in  column  $,,  opposite  third  span. 

Second. — Take  sum  of  coefficients  (multiplied  by  their 
jP's),  opposite  J?,*  to  JP/,  inclusive. 

Knowing  the  maximum  shears  and  bending  moments, 
the  maximum  stresses  are  then  easily  obtained. 


60 


THE  CONTINUOUS  GIRDER. 


1=2.40' 


The  following  is  a  graphical  solution  of  the  same  example. 
Tables  (&)  and  (<")  can  be  filled  by  means  of  graphics,  as 
can  also  Table  («),  by  using  Pro/.  Greene's  Method  of  Area 
Moments,  which  is  fully  explained  in  Part  II  of  Greene's 
Trusses  and  Arches. 

Let  it  be  supposed  that  Table  (ft)  has  been  filled,  either 
by  computation  or  by  Greenes  Method  of  Area  Moments; 
we  have,  then,  the  bending  moments  over  the  supports,  or 
the  "pier  ordinates."  Consider  only  a  single  concentration 
at  the  second  apex  of  the  first  span,  and  call  it  J?*;  then,  if 
the  first  span  were  discontinuous,  the  moment  polygon 
A  c<  IB  would  enable  us  to  determine  the  bending  moments 
at  other  apices  of  the  span;  bat  the  girder  is  not  discon- 
tinuous, and  the  load  Pf  induces  a  negative  moment  over 
the  second  support,  which  may  be  designated  by  IB  J?,,, 
then  the  other  negative  moments  are  given  by  the  ordi- 
nates in  the  triangle  or  polygon  A  B  ]$>.  The  differ- 
ence of  the  ordinates  of  the  two  polygons  determines  the 
magnitude  and  kind  of  bending  moment  at  each  apex  of 
the  first  span.  The  load  jP/  induces  a  positive  moment 
over  the  second  support,  which  may  be  designated  by  C  Cg, 
then  the  ordinates  between  the  lines  1$  C  and  J2.>  C3  will 
give  us  the  moments  at  the  apices  of  the  second  span.  The 


APPLICATIONS.  61 

moments  in  the  third  span  are  given  by  the  ordinates  be- 
tween the  lines  €•>  I)  and  C  ID. 

Having  given  an  outline  of  the  method,  we  will  now  pro- 
ceed to  give  it  in  detail. 

First. — Assume  some  value  for  JP/,  as  unity,  ten  or  one 
hundred. 

Second. — Form  the  stress  diagram,  Fig.  16,  assuming  for 
the  pole  distance  some  value  as  unity,  ten  or  one  hundred. 
Assume  the  pole  in  such  a  position,  that  the  closing  line 
A  1$  to  the  equilibrium  polygon  A  c,  IB  shall  be  horizontal, 
and  construct  the  equilibrium  polygon  A  c,  IB. 

Third. — From  Table  («),  we  find  the  bending  moment 
over  the  second  support  to  be  —  13.54  JPf;  now,  if  IPf=l, 
and  the  pole  distance  ==  J,  then  the  ordinate  IB  J^2:=13.54j 
laid  off  to  the  tcale  of  A  IB.  Draw  A  IBa.  Scale  the  ordinates 
b,  b,,  c,  c*,  etc.,  and  place  the  results  in  Table  (c),  column  3, 
opposite  _P/,  fgj  etc.  The  results  will  be  found  to  agree 
with  those  computed. 

Fourth.— From  Table  (a),  we  find  that  C  C2=  +  3.762 
Pf]  hence,  lay  off  C  Cs=3.76  to  scale  of  A  IB,  and  draw 
1$2  C2  and  Cs  ID  and  fill  out  the  remainder  of  column  3, 
Table  (c),  by  scaling  the  ordinates  between  these  lines  and 
the  horizontal. 

In  like  manner,  each  column  of  Table  (c)  can  be  filled 
in  a  comparatively  short  time. 

To  fill  Table  (ft),  proceed  as  follows: 
Fiist.  — Draw    G  H,  Fig.    16,    parallel   to   A   _R,  then 
72  H     $,  and  F  H=S;. 
Second.— Draw   G  If,  parallel  to  B3  C2,  then  K  Ht=82 

'-&'. 
Third.— Draw  G  H3  parallel  to  C,  1),  then  K  H2=  -& 

=  +  «,'. 

Proceed  in  like  manner  with  each  load. 


62  THE  CONTINUOUS  GIRDER. 

*     EXAMPLE  2  — VARIABLE   L 


K-- 


lt=52,  lf=6fi,  lf=65,  14=52.     8=4.     r=l,  2,  3  and  4. 

m=l,  2,  «•'>',  4  and  £. 
i  t       /  -« 

L=240=I,      L=300=I;i 


1^=420=1, 

It=r>2o=i0 


Wr=6.7 


I5  =300=IS        ltn=360=I3 


It  =420=1IS  I6  = 

I2  =300=1,,  /7  =300=1, 

I3=300='iio  }8  =240=Ir> 

IA  =240=1,  1=300=1 


Determine  the  value  of 
From  (A), 


M2  = 


In  which  the  several  terms  have  the  following  values : 


*    See  page  131,  Alhjem,eine  Theoricund  Berechnung  tier  continuirlic7i.cn  untl 
einfachen  Trager,  by  J.  Jacob  Weyraueh,  Ph.  D. 


APPLICATIONS. 


First  Span 

L=52.        e,  =e, 


, 

A, 

*:  4 

e,  =      10,50 

4-  0.300 

+    6 

e,  =      13  75 

/\,  =    -|-  0-337 

37 

-e,=      3125 

-  0  337 

198 

e4=      31,50 

-  0.300 

—   237 

c,  =      43.50 

+  0.433 

+   685 

f'G  =     46  00 

+  0  289 

541 

e7  =      48.00 

+  0.201 

427 

ea  =      50.00 

A,  -    +  0.243 

+   584 

f  1825 

:  +  -si"    :  +  35-omi 

Second  Span, 


A, 

A'-'/      \» 

A.  <- 

ef 

cv 

A              ^        ^"' 

(3-2  +) 

A                    " 

A.       z 

A,     t 

c,  =      1.50 

A/  = 

-0.256 

73 

2 

0 

<V'=5:       K.OO 

Af-— 

-  0  222 

124 

() 

0 

e*  =     4.75 

A,  = 

-0.311 

267 

20 

1 

l»4'=:       6.75 

-  0.467 

553 

59 

2 

cs  =  19.25 

A,  - 

+  0.467 

+         1285 

+       417 

+      51 

^  =r  23.50 

[0.311 

+         •  972 

391 

+      62 

<?7  5==  42.75 

A?  == 

0.311 

1261 

957 

374 

e,  ==  47.00 

A,-  = 

-  0.467 

1931 

1603 

746 

rv       5625 

AP  — 

+  0467 

+          1968 

+     1876 

+  1278 

r/0^:  58.75 

|  0.311 

1313 

+     1279 

970 

PW==  60.75 

A«== 

-I  0.222 

938 

927 

+    765 

r;,^  62.25 

A/*— 

|  0167 

+           706 

701 

+    620 

Ctt±=  63.50 

A«— 

+  0.166 

+            701 

+      700 

+    657 

+  3674 

+  3641 

+  3280 

+  ^6f7^ 

-f^ 

U     F                      +3644        I   tei 

-)dl=Fa. 

65 

^~f-                         65 

|   50  461 -=!<;. 


64                           THE  CONTINUOUS  GIRDER. 
Third  S/Hfit. 

e,, 

A. 

i:i-d-evy 

,  £ 

>    '  e"  } 

i 
11 

32 
90 

-f       343 
331 
858 
1443 
+     1775 
1202 
865 
+     1004 

^~30~82~ 

,; 

I 

A,     l 

~0 
0 
0 

1 

4 

H-      39 
49 
—    318 
638 
+  1316 
+    972 
+    755 
+    937 

~Tl07 

e,=      1.50 
e»=     2.75 
e,  =      4.25 
e4  =  -.    6.25 
r,  =  :    8.75 
e,=  18.00 
e7  =  22.25 
es=  41.50 
<>9  =  45.75 
elo=  58.25 
e,,=  60.25 
«„=  62.00 
e,3=  63.50 

-0.154 
-0.155 
-0.206 
-0.289 
-  0.433 
:  +  0.433 
f  0.289 
-0.289 
£p=    -0.433 
A«F=  +  0.433 
A//=+  0.289 
A«=  +  0.206 
/\/3=  -f  0.238 

44 

80 
160 

319 
644 
+          1138 
874 
1163 
]  782 
+         1827 
1220 
-870 
1005 

~T742~ 

+  *W_      ,    ,,„.,.     v                     S°M  _      , 

/r 

65 

65 

+  M)7-         4-soo 

-  T 

65 
Fourth  Spun. 

<•„                  A; 

i:i-(l~e,,r 

f?(3-2  A_) 
/ 

A,         L 

*,—         2.00                   0.110                      33 
e..,  =         4.00      A,   ~-  --    0.095                       55 
f,=         600                   0134                    111 
CA  -           8.50     A4   :       0-200                    224 
(V=          17.50                   0.143        +          274 
0;  -         20.75      A«   =       0.151                    320 
e,  ^       38.25                   0.151                    101 
e^=        41.50     AS   =      0.143                   384 

1 
4 
13 
39 
+        102 
143 
337 
346 

-614 

—  495 

±*»             22807    F                        d** 

- 

E\o)                                                              *                                              >-*& 

o&                                                                  o& 

'°           4  • 

APPLICATIONS.  65 

Collecting  the  values  of  F,  we  have 

.F,"=  -f  35.096 
-    Fa=  +  56.528         FS=--  -f-  56.061         F*  =-- +  50  Jfil 

Fs=  +  4%315         F,'=:  +  -47.^5         F/=  -f  47.£00 

#=    -  ^.£07         JFV  =          9.519 

From  (e), 

0,  =  6  E  It  =  520  E '  (by  making  L  '=6  E) 
Og  =  6  E  It  =  560  E'  (by  making  E'=6  E) 

03  =  6  E  It  =  520  E   (by  making  E'=6  E) 

04  =  6  E  I  =  240  E '  (by  making  E  '=6  E) 

From  (mj, 
pa=0s  (l,^F.;)=520  (121.061)  E'=      62,951.7  E' 

^=04(l8-^rF;)=240  (112.415)  E'= 26,979.6  E' 

From  (n), 
«— ^(t.+A^+ACt+R) 

=  560  (87.096)  E'  +520  (121.523)E'=  .    .    111,965.7  E' 

&=M4+^;)  +  M^+^) 

=  520  (115.461)  E'+560  (107.215)  E "=  120,080.1  E' 

/^M^  +  ^')-M<(^  +  ^) 

=  240  (112.800)  E'+520  (40.193}  E'=  47,972.3  E' 

From  (o), 
&=V*  (ls-\-Fs)=520  (121.061}  E  =       62,951.7  E' 

ft=0,  (13+F;)=560  (112.415)  E  =       62,952.4  E' 

fc=03  (14+ F^)=520  (42.481)  E= 22,090.1  E' 

Since  kr=  ~,  I  Pr  lr  (l—kr)=lJ  Pr  (lr—ar).     Now,   we 
ir 

lr 

have,    from    (&),    for     a    uniform  load,    -   Pr=  lwrdar, 

0 

lr 

:•.    ^'  Pr  lr  (l—kr)-  =    wf  d  a,.  (lr~ar)==      -  wr  //,  and   we 


66  THE  CONTINUOUS  GIRDER. 

can  write 

r  P,  i,  (i-k,)=  4- ».  v=  4-  *M»^=  •  • 


v  P2  I,  (l-k»)= 


Also, 

ev 

IPr(er— ary=   Cwrdar(e,:—ar)3=   -j-  wr 
0 


^  Pf  (e-ary=--      wr  d  ar  (e^ar)9=--  --  wr 


0 

Substituting  these  values  in  (/)  and  (r/),  they  become 


v=lr 


v=l 


4k  lr  3 

v=lr 

v=l 

'     3  wr  (  et__  \ 

~T~1  lr      I 

9=4 


APPLICATIONS. 
VALUES  OF  H  AND  H'. 


67 


4 

4 

Is 

i. 

^P. 

QS 

l~ 

V 

, 

•4 

-, 

i 

*.u 

\ 

\  ^ 

'  ' 

**  ~° 

V 

^  I 

V 

<\ 

<r 

<f 

<f 

<• 

<f 

1 

-f-   70 

3 

0 

2 

0 

3 

2 

-f-  232 

23 

0 

12 

0 

23 

3 

-  6180 

126 

2 

60 

1 

106 

4 

-  8173 

530 

15 

262 

7 

431 

5 

-f-  29815 

-f-  10365 

+  986 

1043 

39 

4   2292 

6 

+  24884 

+  H767 

-f-  1459 

4-   8003 

-|-  699 

3781 

7 

-f  20514 

49249 

-  15980 

4   9464 

1090 

15153 

8 

-f  29207 

88767 

—  35059 

43059 

-13188 

16412 

9 

4  117518 

+  71927 

78303 

-  29183 

10 

+  81258 

+  57000 

4-  112239 

4  76693 

11 

H-  59538 

4-46518 

4  77063 

+  58589 

12 

+  45396 

-f  38580 

4  55895 

-f  46830 

13 

-f  45445 

4-  41523 

4-  65156 

4  59533 

//; 

5 

H; 

H, 

q 

k 

-67777  w,  +58147- WM  -W 

.-.  H2=  4  389,584.9. 
H3=  4-  112,794.0. 
JZ,^  .-  43,643.8. 


g  +51270  wa  -150762  w3  -6514  > 

HJ=  454,105.9. 

H2=     -  1,039,860.1. 
H3=  —     331,676.4. 


From  (h), 

^^  p;   vpais  (1—k,)  ot= 

-56.06 (14,153.7) 520E=   -  412,597,339 E' 

<j)    XT'"     \'Z57     //         I-    \    ft  

—    **    ™  1     ""*    •*  /    "1    V  •* — ~"'i )    ''2 

-  2-35,09  (9,058.4)  560  E'=  -  356,002,336  E' 
4-  H,  fli==+  389,584.9  (520)  £'=  4  202,584,148  E' 


I 


-H;  o,=  +454,105.9  (560)  E'=  -f  254,299,304  E' 

-  311,716,^23  E' 


68  THE  CONTINUOUS  GIRDER. 

+     V   v  P    7    /  7      I-  \  0 
-*  3     — '    *  J    kg    V  •* "'S )    U2 

-47.41  (4647.5)  560  E  ^     -  123,389,266  E' 

| 2  F"  -  P9  I  (1 k. )  Q.= 

X*=  -  2  (50.46)  (1 1153.7)  520  E '-     -  742,763,530  E' 

X;+X;     +#,  0,=  + 112,794.0  (560)  E,=  +    63,164,640  £' 

-Hi  #.,=+1,039,860.1  (520) E'-^-  +  540,727,252  J5' 

-  262,260^904  ^r 

-  F4IP4l4(l-k4)  o,= 

+9.52  (9055.4)  520  E'=  -  +   44,842,703  E' 

-  2  F;  ?  P3  I,  (l—ks)  0,= 

-47.80  (4647.5)  240  E'=         106,632,240  E' 

+  Ht  0a=  --  43,643.8  (520)  E^          22,694,776  E' 
~  H;  o,=  +  331,676.4  (240)  E'=-.  +    79,602,336  E' 

'  4,881,977  A' 

From  (75),  Jr= —  wr  I?  Or_t. 

4 

From  (77),  Br=  -  wr  I?  <>r+). 

Therefore,  A^=         4~  w2  1°  0,=  -239,198,375  E' 
4 

A,=          4-  w,  II  o.=  -     84,584,500  E 
4 

A4=  -     4-  w*  V  °f=    -  122,469,568  E 

4 

B<=  -  w,  If  02=    -  131,890,304  E 

B.=         4-  wt  U  o,=  —  239,198,375  E 

4 

B,= —  w3  If  0A=  —    36,250,500  E 


APPLICATIONS.  69 

From  (p),  since  ct=0.  and  ca=l, 


C3  - 


—  2c4fc—csK  _      ~  2,590,360  E' 

~  ~ 


From  (>),  since  <i,~tf,  and  ds=l, 


—  2  dA  fa—ds  ft,          —  1,109,880  E' 

d5= -y> -  ,„ 

Therefore, 

(   (  —239,198,375  E'—  210,013,191  E'  ~\ 

—  101,703,032  E'     -oo- 
'   '         5.38o 

-311,716,223^'  J 
f  —  84,584,500  E'—  60,224,626  S' —239,198,375*.' ^  (—1.524) 


^  __      ~  1      j 


=Tl09.HHOJP' 


—  202,036,278  £' 

—  262,260,904  E' 

—  122,469.568  E  +  22,147,927  E '  —  36,250,500  E ' 

—  27,029,904  E' 


—      4,881,977  E' 


—  5,385  (2.33) 


—  2,590,360  E 


—  (—  131,890,304  E') 


M.= 


(-  2,966,675,: 
!_  ™^™™i         12.565 


!  +    893'13°'719     +  (-131,890,304) 

I-    163,602,045  J 


M2=  -2015.66  —  639.68=    -2656. 

The  moments,  produced  by  the  loads  in   the  respective 
spans,  can  be  easily  deduced,  as  follows  : 

First  span  loaded— 

M  —        ~     u  '~*3  '  2       R  _L  ~  V"    1 

~  (c.  ft.)  ft,  ft,       '•  TdJfT)  * 

>  (-101,703,032)  (5,385) 


|  ,          -,,  ,  _  4,)4 

M,=  -  640 


70  THE  CONTINUOUS  GIRDER. 

Second  span  loaded— 

M,= 


,.- !  ((—  239,198,375  —  210,013,191)  (5,385)       )_  _    1  574 

*  **  1,109,880  \(_  202,036,278  -  239,198.375)  (-  1,524)  ) 

Third  span  loaded— 

>.\ 


-  1          C  (-  84,584,500  -  60,224,626)  (—  1  ,524)  i  I    1  4.O 

2=    1,109,880  j  (_  27,029,904  -  36,250,500)  J          ....... 

Fourth  span  loaded— 


-       __ 

1,109,880  j  _j_  22,147,927  \  Sum-    -2656 

THE  SAME  EXAMPLE  WITH  I  CONSTANT. 

Fr-* 

c5  ^ 

' 


,=0,  ^,=1,  d,=  —  3.6,  dt=  -f 


2901  () 


,=  --  r  ws  I,  A,=  --  r  v.,  //          ^1,=  —  - 

444 

<=—  4-  w/  ^*       Br=^-  ^-  w,  //        ^.,= 

-y  4  4 

Then, 


4 


-  4-  ™< l?  (  13-4) 

4 


APPLICATIONS. 
f  —  919,993.7  w  -f  247,162.5  w 


71 


--J  -h  247,162.5  v  -      68,656.2  v  -  2587 

-  35,152,0  w  —  471,036.8  w 
The  partial  moments  are  easily  obtained,  as  follows : 

First  span  loaded— 


M- 

~ 


2901.6 


(—471,036.8  w)  =-. —  1088 


Second  span  loaded— 

1 


w  +  247,162.5  w)=  ...  —  1554 

Third  span  loaded— 
t=5IKfi-  (247,162.5  v  —  68,656.2  v)=  .    .  ±  136 


Fourth  span  loaded— 
(-  $6,126.0  »)= 


Sum  --=  —  2587 


Spans  loaded. 

Variable  I. 

Constant  I. 

Difference. 

Per  cent,  of 
Variable 
I  Results. 

FIRST. 

—  1134 

-1088 

-46 

.041 

SECOND. 

-  1574 

-1554 

-20 

.012 

THIRD. 

+    142 

-f    136 

+    6 

.043 

FOURTH. 

90 

81 

9 

.100 

ALL. 

-2656 

-2587 

-  69 

.026 

72  THE  CONTINUOUS  GIRDER. 

EXAMPLE  3.— THE  SABULA  DRAW. 

We  now  propose  to  compare  the  bending  moments  in  the 
Sabula  Draw,  considering  the  moment  of  inertia  as  constant 
and  variable,  respectively. 


Variable  moment  of  Inertia — 

From  Fig.  20,  is  obtained  the  cross  section  of  each  mem- 


APPLICATIONS.  73 

ber,  and  the  length  of  each  vertical;  the  moments  of  inertia 
for  each  section  are  readily  deduced  as  follows:  neglecting 
the  moment  of  inertia  of  the  section  of  the  member  about 
its  own  axis,  it  being  very  small,  in  comparison  with  the 
moment  of  inertia  of  the  truss  section. 


£=(17.7+17.6)  (12.5  y-  144=  40=  £ 

£=(29.1+22.5)  (13.4  )2-144=  60=  £ 

£=(29.1+28.5)  (14.25)2--144=  80=  £ 

£=(34.5+31.5)  (15.15)^144=110=  £ 

£=(34.5+27.5)  (16.05)2— 144=110=  £ 

£=(33.8+275)  (16.9  /-144=120=  £ 

£=(31.5+28.5)  (17.8  )2-:-144=130=  £ 

£=(47.5+36.0)  (187  )2--114=200=  £ 

£=(47.5+48.0)  (200  )2--144+270=  £ 

FIRST  SPAN. 


=104.5 


c4=  75.5 
^,=113.5 
^=132.5 
V=151.5 


From  (124),  we  have, 
Ma--     z4-=-(- 


In  which, 

X:=H2  o,  —  F2  I  P2 1,  (1  —  h)  (>s 
X,"=    -  H,  02  —  2  Ft"  £  P,  1,  (1—k,)  0, 

v-=l 

«      (  4  1 3 

v=L 


THK  CovriNrors  UIKDKK. 


v=lt 
;  £/,(/,-/•:  )+6  Kl(L-Fs) 


9=1 

*      f3 

IT"  v          A         r 

'*        -      A  -IT 

r     I, 


v=l 
Fs=  Z 

•     I  4 


/_,  1  Ir-,  I 

We  will  first  compute  those  co-efficients  that  do  not  depend 
upon  the  loading.     For  a  uniform  load  over  all, 

t-=/ 

H!=  I  4-3  4 

r  4  (  «S      ) 


H,=  r 


9=1 

•'  >'~ 


APPLICATIONS. 


75 


Ft  fat  san— 


<?/  == 

28.5 

A,  =  -f 

2.250 

-  -      1,484,437 

es  = 

47.5 

A*  ~  + 

1.125 

5,726,996 

e3  = 

66.5 

A*  =  + 

0.921 

18,011,347 

e.  = 

104.5 

A*  = 

0.204 

24,327,379 

e,  = 

123.5 

A*  =  -r 

0.173 

40,245,185 

efi  = 

142.5 

A«  =  -f 

0.727 

-  299.773,934 

e7  = 

161.5 

Ar=  + 

0.350 

-  238,099,317 

Multiplying  this  sum  by 

H;. 

Second  span  — 


3wt 


—  627,668,595 
obtain  —  2,615,286 


, 

L\  <>               ^ 

e<  = 

18.5 

/\*  =  — 

0.052 

1,215 

es  = 

37.5 

A?  =  — 

0.107 

19,043 

es  = 

56.5 

As  —  — 

0026 

14,342 

e4  = 

75.5 

Ax  —  — 

0.030 

35,397 

e,  = 

113.5 

A,-- 

0.137 

422,318 

C6  = 

132.5 

A«-=  — 

0.166 

691,825 

e7  = 

151.5 

A;-- 

0.333 

-  1,709,655 

-  2,893,795 


Multiplying  this  sum  by  ^-,  we  obtain  — 7 

4 

First  span— la=180- 

e<=  28.5     A^  +  2.250  A/ 

«•,=  47.5     A»='+ 1-125  A» 

e.^  66.5     A.?—  +  0.921  A. 

c.=  104.5    A,—  -f  0.204  Ax 

*,=123.5     A«=  -f  ai73  A, 
«—  -f  0.727 


u-.,~H* 


=  -h 


52,085 
120,568 
270,847 
232,797 
325,871 


6^=142.5 

«7=-161.5     Ar=  4-  0.350 


AX-  4- 

A,  ^—  + 

A*  <S=  -f  2,103,676 

A;  «J=  +  1,474,299 


/f=:3*2,400     )  4,580,143  (  141.3(5     /•'  . 


70  THE  CONTINUOUS  GIRDER. 

Second  span— L= 180  . 

a  ,                                      i' 

{  '••*    &>•    's •""     ('!•)     I  \  'l  '2           V'2         '17       I,  /  \ 

<?,=  18.5               -0052-        8,951  84,225 

e,      37.5               -0.107-      69,967  314,404 

e~_     56.5               —0.026-      35,440  102,657 

r/      75,5     £4=  —0.030-      66,521  140,725 

r,     113.5                   0.137—   552,403  -    758,695 

r,,     132.5     A«-    -0.166—    801,442  -    950,321 

c,     151.5               —0.333—1,813,227  -1,936.284 

£=32,400  ) -3,347,951     /!^-32,400  )  —4,287,311 


—  103.33=*V  —  132.32=*; 

Next  find  the  values  of  X*  and  X,',  for  a  uniform  load 
over  all. 

o,=.27(t  E',  (ty—40  E',  in  which  E'  —~n  K. 
+  H.  o,=    -  723,449  (270)  E'  w.  =          -  195,331,230  w,  K' 

F,  .    103.33 

--  j-   w*  1:  <>,=  H  --  -  - 

(32,400)  (270)  £"  MJ,     =  +  451,965,420  ws  E' 

+  256,634,190  IP,  A1' 

_;/  VA,  //?/  /*=  -141.36  (32,400)(40)w,  £'=    -  183,202,560  wt  E  ' 
-H,  os=  -f  2,615,286  (40)  wt  E'=  +  104,611,440  wt  E' 

78,591,120^  E' 
&=40  (180+141.36)  £'+270(180—  132.32)  E'=  +  25,728  E'. 


A»=  —      ~  "'"  ^'  270  £"=  —  393,660,000  E'  w,. 

4 

B,=  --  4-  ^  //  40  E'=  —  58,320,000  E'  w,. 
4 

Then, 

(  —  393,660,000  w,E'} 
1        \  —    58,320,000  w,  E  '  \ 
"     -  5T4M7?  78,59i;i20  «,;  £'  "<lf  ^» 

H   256,634,190  t»3  B{ 


APPLICATIONS. 


77 


Jfirst  spun  alone  loaded— 

f  —    58,320,000  w/i 
±-    78^91,120  W| 


I  —  137,000,000  w,  } 

Second  span  alone  loaded— 

(  —  393,660,000  w,} 
M^  +  256,834,190  w,  I 

-  137,000,000  ws  J 

EXAMPLE  3o. 

This  is  the  same  as  Example  3,  with  the  moment  of  inertia 
considered  as  constant. 

From  (142), 


720 


A,= w.  11=  —  145,800  w,. 

4 

B,= 7-  w<  lf--=  —  145,800  w,. 

4 

Therefore, 
Ma  -        ~        r  ™       and  if  w,=w,F=w1  Ms=    -  4050  w. 


No.  of 

Loaded 
Span. 

FIRST. 


Bending  Moment 
Variable  1. 


-  2661  w, 


SECOND.       —  2661  w. 


BOTH.          —  5322  w 


M2 

Bending  Moment 
Constant  I. 


-  2025  w, 


—  2025  w 


—  4050  w 


Difference. 


636  w, 


636  w 


1272  w 


7-S  THE  CONTINUOUS  GIKDEK. 

EXAMPLE  4.— CONCENTRATED  LOADS. 

Let  us  consider  the  Sabula  Draw  again,  but  with  single 
concentrated  loads,  instead  of  uniform  loads.  Supports 
level.  a,=»7.  o.?— /?•>. 

See  pages  73  and  74  for  the  value  of  M,. 

From  Example  3  : 

#=40  (180+141.36)  #'+270  (180—132.32)  £"=  +  25,728  E'. 
Ft"=--  +  141.36.     #=103.33.     F,=    -132.32. 
^=270  E'.     ^=40  E'. 

The  values  of  H3  and  Hi  depend  upon  the  position  of 
the  loads.  Let  us  take  a  load  in  the  first  span  £7'  from  the 
left  support,  and  one  in  the  second  span  123'  from  the 
left  support;  then  we  have  for  Hi  and  H,,  the  following: 

First  span-     l,=180'.    a,==£7'. 

V=l 


*=  28.5        e—a<. 

e,=  47.5 

es=  66.5      +  9.5  A*=  +  0.921  88  P, 

e4= 104.5  +  47.5  ,  =  +  0.204  680  Pt 

+  66.5  A  5= +  0.173  —  1,292  P, 

+  85.5  A<r=  +  0.727  -  10,097  P, 

+  104.5  t\!=  +  0.350  —  8,068  P, 


-  20,225  P,= 
NOTE.—  er  must  always  be  greater  than  or  equal  to  a,. 

Second  span-     1=180'.    ax=123'. 

v=l 
H,=  I      A    |JjPg  (^— a>I  ^  t  -ii— ^  J  P,  (c,—a2Y  | 


APPLICATIONS.  79 

e,  -  18.5 

e,  =  37.5       e—a, 

e3  -  56.5 

ej,  -  75.5 

es  ==  113.5 

e,  ==  132.5     4-    9.5     A5  =     -0.166  13  P 

e7  =  =  151.5     +  28.5  =  —  0.333        —  172  P2 


—  185  P,=H, 
NOTE. — e,  must  always  be  greater  than  or  equal  to  as. 

r  -f  #,  0/=    -  185  (270)  E'  P2    •  49,950  E'  Pg 

*  -          -  F:  P,  (ls-n,)  (),= 

(  H-  103.33  (57)  (270)  E'  P,=  +  1,590,248  E'  P3 


98  E'  P2 
(  —  H;  03=  +  20,225  P;  (40)  E'=  +     809,000  E'  P, 

(  -  141.36  (123)  (80)  E'  P<=     -  1,390,928  E'  P, 

~581~,928  Y'  P; 

From  (i)  and  (J), 

/ 

A  p    12  ( <r>  i,       &  Z-2  i    £.3\  ft    T?r    T 

./lo JL    •?    t">       \  &     A/ d     A/    — — /V      I      C/      Xl/         _//. 

Or,  by  Table  I, 

A,=    -  P  //  (0.285,144)  270  E'  -     -  2,494,269  Ef  P,. 
^/=    -  P^/ (0.285,144)    40£"=  369,520  E' P,. 

Therefore, 

f— 2,494,269  ^'  P/l 

M  _  I        j  +  1,540/298  E '  P,  {  __  j  -18  5  P2  \ 

~  51456  E'}  -       369,520  E'  P,  f  ~~  1  —18.5  P,  } 
581,928  E'  P; 

Or,  if  P;=P,? 
3f3^    -  37.0  P. 

First  span  alone  loaded— 


(SO  THE  CONTINUOUS  GIRDED. 

Or, 

»51,448/>=18.5/>, 


'       556  , 

Second  span  alone  loaded— 


Or, 


-  953,961  J>,     =  18.6  P, 


Since  IP,  and  IP,  are  symmetrical  about  the  center  of  the 
draw,  the  moments  over  the  center  pier  should  be  equal, 
as  shown  above.  If  our  work  was  absolutely  correct,  and 
decimals  had  been  used  to  several  places,  the  quantities 
in  the  parentheses  above  would  have  been  the  same  ;  as  it 
is,  the  quotients  are  practically  equal. 

*    EXAMPLE  4  BY  GRAPHICS. 

For  any  load  in  the  first  span  of  a  two  span  girder,  we 
have,  from  Greene's  Trusses  and  Arches,  Part  //, 


EI  El  El 

i 
In  which: 

distance  from  the  left  support  to  the  ordinate  y, 
of  the  equilibrium  polygon  A  C  IB.  Fig.  21,  page  72. 

ordinate  of  the  polygon  A  C  IB. 

distance  from  the  left  support  to  the  ordinate  y'f  of 
the  polygon  A  IB  IB,. 

ordinate  of  the  polygon  A  IB  IB,. 

distance  from  the  right  support  to  any  ordinate 
a  of  a  polygon  in  the  second  span,  similar  to  AS  _Z?,. 


*    For  this  excellent  graphical  method,  the  author  is   indebted  to  R.  H. 
Brown,  C.  E.,  First  Assistant  Engineer  of  Boston  Bridge  Works. 


APPLICATIONS.  81 

ordinate  of  the  above  polygon. 

jT— The  moment  of  inertia  of  the  cross  section  of  the  girder 
at  any  ordinate  y',. 

2 

J— The  moment  of  inertia  of  the  cross  section  of  the  girder 
at  any  ordinate  y's. 

_Z?— The  modulus  of  elasticity. 

/  2 

Since  l,=L=l,  and  1=1=1,  we   have,   considering    JE  as 
constant, 


Now,   -  '— ^  equals  an  area  multiplied  by  the  distance 

of  its  center  of  gravity  from  the  left  support,  and  hence  we 

or   ?/ 
may  write  2  — '-7^    =  A  a,  in  which  A  represents  the  area 

and  a  the  c.  g.  distance. 

In  like  manner,  #  -  ''' .  '  =2  B  b.  Therefore,  we  may 
write,  ^1  a— 2  B  b. 

With  any  scale,  lay  off  the  horizontal  line  A  12,  Fig.  21, 
and  divide  it  into  panel  points  at  ?>,  r,  d,  etc. 

With  any  scale,  preferably  a  large  scale,  lay  off'  a  load 
line  A  ID,  equal  unity,  and  assume  If  also  equal  unity, 
then,  assuming  the  pole  in  such  a  position  that  its  closing 
line  will  be  horizontal,  construct  the  equilibrium  polygon 
A  C  IB  and  scale  its  ordinates  (the  lengths  are  given  in 
Fig.  21.) 

Not  knowing  the  proper  position  of  the  closing  line,  let 
us  assume  a  position  as  A  12»,  making  B  B2=12.89  and 
scale  the  ordinates  b  b3,  c  C2J  etc. 

Divide  each  ordinate  of  the  polygon  A  C  B  by  its  proper 
I  (given  in  per  cent,  of  the  center  I  above  Fig.  21 ),  and  lay 


82  THE  CONTINUOUS  GIRDER. 

off  the  results  downward  from  A  It,  forming  the  polygon 
Ab5c5.    .    .    .  B.      The  ordinates   are    80.00,   118.18, 

etc.,  as  shown  in  Figure  21,  page  72. 

Now,  find  the  area  of  this  figure,  either  by  computation 
or  the  planimeter,  and  also  its  center  of  gravity.  The  center 
of  gravity  is  readily  found  by  cutting  the  polygon  out  of 
stiff  card  board  and  balancing  it  upon  a  needle  point. 

The  area  =  109 15.1= A,  and  a=nr>f>. 

Proceed  in  like  manner  with  the  polygon  A  B  B,,  de- 
ducing the  figure  A  b-;  c;;  ....  />,,  which  has  an  area  of 
;>.'f(>!t.(>l=-B',  and  V =100.8. 

Let  B  B,  represent  the  true  magnitude  of  the  pier  ordi- 
nate  //„,  then  from  the  triangles  A.  B  B,  and  A  J>  Bn  we 
have, 

y,',:ylt  '.'.c  c2:e  c,,  or,  c  e,=      r  e  e2.      Then, 

x ,(€€,)          ?/„    v  x2(ec.^  y^       ,     , 

—      f —  r     —     -   — j ,     UI ,/)!/—  r     1J      v   . 

I  Vo  I  'Un 

-,  v,>    , , ,  j ,  (Ad)  '»/,', 

Hence,  A  a=2  -~-   II'  b',  and  ya  =     ^  ^/^  , 

Which  becomes 

10915.1X65.5X12.8 
•2X2469.61X100.8 

Since  H=l  and    Pf=l,   the  bending  moment  over  the 
center  pier  is  —  18.38  JP*.     We  obtained,  by  computation, 
-  18.3  JP;V,  which  shows  the  graphical  method  to  be  accu- 
rate enough  for  all  practical  purposes. 

It  would  have  been  more  correct  to  have  taken  the  ordi- 
nates y  at  the  points  where  the  values  of  J  change,  but  the 
result  would  have  been  but  little  different. 

The  above  computations  can  be  very  readily  made  by 
means  of  Thatcher's  Calculating  Instrument. 


APPLICATIONS. 


83 


A.      rf 


Fig.  17,  page  72,  and  Fig.  18  are  diagrams  for  the  Sabula 
Draw,  with  a  concentration  at  each  apex  in  both  spans. 

A  a—B  b.  or  A  a--=  -^4-  B'  6.     Therefore, 
V* 

_(Aa)y0          65568 X  148x94.6 
B'  b'  36865X2045 

Since  H—fif^  the  bending  moment  over  the 

pier  is  121,800x50-= 6,090,000 

Considering   /   as    constant,    the    bending 

moment  is 4,732,000 

7^58~000:= Diff. 


APPENDIX. 


*  The  general  equation  of  the  elastic  line  can  be  deduced 
as  follows : 

Vertical  forces  acting  upon  a  girder  cause  a  change  of 
shape,  lengthening  the  originally  parallel  fibres  on  one  side, 
and  shortening  or  compressing  them  on  the  other.  Between 
the  lengthened  and  shortened  fibres  there  is  a  plane  which 
undergoes  no  change  in  length ;  the  centre  line  of  this  plane 
is  called  the  neutral  axis  or  the  elastic  line. 

Thus,  in  Fig.  (<r),  m  o  is  the  neutral  axis,  the  fibres 
above  being  compressed,  and  those  below  lengthened.  Upon 
the  three  following  hypotheses  we  shall  deduce  the  equation 
of  the  elastic  line. 

I.  All  planes  perpendicular  to  the  axis  before  the  bend- 
ing or  flexure,  preserve,  during  the  bending,  their  perpen- 
dicularity and  their  forms  as  planes. 

II.  The  change  in  length  of  a  body  subjected  to  a  force, 
is,  within  certain  limits,  called  the  elastic  limits,  propor- 
tional to  the  intensity  of  the  force. 


See  Merrimaii's  Theory  and  Calculation  of  Continuous  Beams. 


86  APPENDIX. 

III.  The  change  of  shape  is  so  small  that  the  length  of 
the  neutral  axis  is  sensibly  the  same  as  its  horizontal  pro- 
jection. 

In  Fig.  (a),  we  have  a  longitudinal  section  of  a  portion 
of  a  bent  beam ;  the  two  planes  (i  1)  and  (I  e  origininally 
parallel,  remaining  perpendicular  to  the  neutral  axis  or 
elastic  line  in  0,  and  intersecting  in  c  the  centre  of  curva- 
ture. Hence,  drawing  /  y  parallel  to  a  b  through  o,  the 
lines /f/,  (/  e,  etc.,  denote  the  elongation  of  the  fibres,  and 
we  see,  from  the  figure,  that 

od:od'::df:d'  f (22a) 

or  the  change  of  length  in  the  fibres  is  proportional  to  their 
distance  from  the  neutral  axis.  This  is  a  consequence  of  the 
first  hypothesis. 

Designating  by  If  and  If  the  force  acting  in  the  fibres 
(If  and  <l'  f ,  the  second  hypothesis  gives 

H:H'::df:d'f (226) 

Combining  (22a)  and  (226),  we  have 
H:H'::od:od' (22r) 

or,  the  horizontal  forces  are  directly  proportional  to  their 
distances  from  the  neutral  axis. 

Denote  the  distance  of  any  fibre  from  the  neutral  axis  by 
2,  the  stress  in  it  by  II',  the  distance  of  the  remotest  fibre 
by  e,  and  its  stress  by  H ;  then,  from  (22r),  we  obtain 

Hf:H::z:erQT,Hf=-~- (22rZ) 

Thus  far  the  cross-section  of  the  fibres  has  been  considered 
as  unity.  If  the  actual  area  is  ft-,  the  force  is  L— .  Each 

of  these  forces  If  tend  to  turn  the  beam  around  o  with  a 
lever  arm  o  d'  or  2,  hence  the  moment  of  the  force  is 

H  a  z  II  a  z3  ~    ,,   ., 

•' X*  =  — ,  and  the  sum  of  all  the  moments  is 


APPENDIX.  87 

Mt=  —   Zazr  ..................    (22c) 

c 

Mx  meaning  the  bending  moment  at  any  section  xr  of  the 
beam  in  the  span  lr. 

Since  -  ft  &  is  the  expression  for  the  moment  of  inertia 
of  the  section  ft  ?>,  (22r;)  becomes 

Mx=—  .....................   (22/) 

(5 

or  the  moment  of  the  internal  forces  equals  the  stress  in  the 
remotest  fibre  times  the  moment  of  inertia  of  the  section 
divided  by  the  distance  of  the  remotest  fibre  from  the  neutral 
axis. 

The  line  d  f  denotes  the  change  of  length  in  the  fire 
a  d,  due  to  the  force  JZ,  hence,  if  J£  be  the  co-efficient  of 
elasticity, 
ad:df::E:H   ..................    (220) 

Designating  the  radius  c  o  by  ?-,.,  we  have,  from  similar 
figures,  o  rl/'and  c  ft  d.     (-in  o=a  d). 
a  d:df::r,.:e    ...................  (22  h) 

H          E  Hr, 

-       :         or'  c=  - 


Substituting  this  value  of  e  in  (22/),  we  have 


r.- 

The  radius  of  curvature  of  any  plane  curve,  whose  length 
is  tt,  and  co  ordinates  xr  and  yn  is 

d  K 
7r      d  x,.  d*  yf  '    ' 

According  to  the  third  hypothesis,  d  u,.=-d  xn  and  (2 
becomes 

!-,.=  'i^-  •    (22m) 


88  APPENDIX. 

Substituting  (22/n)  in  (22A;),  it  becomes 


,.,- 

E  Ix 

Which  is  the  differential  equation  of  the  elastic  line,  appli- 
cable to  all  bodies  subjected  to  flexure  which  fulfill  the  con- 
ditions imposed  by  the  third  hypothesis.  The  values  of  J£ 
and  I  may  be  different  for  each  and  every  section. 


If  (22)  be  integrated,  it  becomes 

•''/•    /• 
;//.,  1    CM*  <l  »V  .  i    « 


o 


If  xr=0,  then  C  =  =  tn  and  we  have 


o         A* 
Integrating  again,  (24)  becomes 


If  xr=6>,  then  C'^=hn  and  we  have 


Xr      r  Xr 

1    rMr  d  xf  c 

yr=hr+tr  xr  +  -g-Jr-  -  J 

o         T*       o 


(23) 


^-  =  tf+-    r  I  ~^  -  (24) 


APPENDIX. 


89 


By  examining  Fig.  2,  which  represents  a  continuous  gir- 
der having  a  variable  cross-section,  and  consequently  a 
variable  moment  of  inertia,  the  following  integration  will 
be  clearly  understood : 

Xr      r 

*  T  4        4--         c    C  Mxd  xr 

*  Integration  01    I 

„    i. 


r   r  .,  e2 

/Mxd  xr       1    /  ,  r   ,  1    c\Jr  j 

-  =  —  \  Mx  d  xr  +  —  I  Mx  d  xr  .    .    . 
T  T    J  T    J 

o        lx  lo  o  Jiet 

x 

1        /•     r 

.    .    .    +      -\MX  d 

iA 


xr 


ex 


But, 


1    f*   r                   1    r*   r                   1    f*   T 
—  I  Mx  d  xr  =  — —  I  Mx  d  xr —  I  Mx  d  xr  .    .    .    . 

7  T  T 

Therefore,  we  can  write  for  (27), 

*    See  Weyrauch's  Continuirlichen  und  Einfachen  Trager,  p.  168. 


(27) 


(28) 


90  APPENDIX. 

xr  e,  e, 

C—MX  d    x,.--      4-  (*Mxdxr-\-  4-  faL  d  x,.... 

I  i*  'i-i          *<o 

e* 

..+  -4-  CjLfx  d'xr 
I     " 

-lx—l    n 


—  (  Mx  d  xr CMX  d  x,.  .    .    .  -     —(M,d  x. 

j  J  J  <J  T  J 

2i  o  o  *  o 


0 

Which  reduces  to 

/>*  or, 

&r 


xr 
-f  —  CMxd  x (29) 

i A 


.    •    (30) 

*     T'->  T>          „ 

v=xr  o 

Substituting  (30)  in  (26),  it  reduces  to 

xr          xr 
yr  =  b  +  tr  xr  +  -jr;  I  —§d  xr  §MX  d  xr 

T^o  o 

v=l  xr  er 

•  •  <3i) 


r  T 

v=xr  v 


From  8,  we  have  Mx^Mr^Srxr-l'  Pr  (xr—ar)  .    .  a<x  .    .  (8) 


APPENDIX,  91 

Hence, 

•''',• 

| V,  fl  xr=Mf  x,.  +  ~  $,. 4  -    ^ l\  ( x~a,y . . .  a<x . . .  (32) 
o 

And, 
xr         xf 

fd-  xr  p£  d  xr  =     ~Mr  4  +  -^   S,  x: 
o          o 

—  Z  pr  <  r,.—ar)»  .    .  a<x  .    ,   (33) 

Now, 

xr          e,.  er          c, 


j'd  x,  j*Mx  dx,.=  j*d  x  ^Mx  d  x, 

xr 

Cdxr      Mxdx, (34) 

•J          */ 


00  00 

xr          t 


e,.          o 
But, 


-^  Z  Pr  (ev-a,Y  .    .    .  «<«  .    .    .    (35) 

Hence, 

er          e,. 

Cd  xr  (*Mr  d  xr  =  4-  Mr  ef.  +  —  Sr  e], 
J          J  2  6* 

o  o 

-|-  JPP  (e,  —  ar)a  .    .    .  a<e  .    .    .  (36) 

x,. 
ldxr=xr—ev, 


92  APPENDIX. 

Therefore, 

xr          c,. 

Cd  xr  CM,  d  xr  --    4-  (xr-e.)  1  2  M  c,   s,  <: 

.  /       ./  ~  ^ 

<?,,          o 

-ZP,(ev-a>)'}     .....  (37) 

Substituting  (33),  (34),  (36)  and  (37)  in  (31),  it  becomes 

yr=hr  +  tr  xr+  \  -I  .I/,.  **+Sr  x*-l'  Pf  (xr-0r?\ 

6  El,   ( 
v=l 


4-     --7-  ? 

6EI>V=*r  J~  T* 

+Srel-l'Pr(er-a,.Y-]t    .    .    .(38) 


Which  reduces  to 

ir=hr-\-tr  xr+    ——  \3  Mr  aj-f-  S,  x*.— I  Pr  (xr—ar]a  1 
6EIn  ( 

+  ±  ?'  [i  -  i- 

6EJX    v=xr   ^/r_;         /,. 

-  I  P,.  (e-ar)*-i  (x-e,.}  1'  P,  (r-n,.y  [-'...;  (39) 


If  we  make  xr-—lr,  then  yr=hr+1,  ef  —  et<  and  Ix=Ii. 
Let  —r  —  .  then,  from  (39),  we  have 


r+l  =  h,  +  tr  lr+  -  \8  Mr  tf-f  S,  lf-*S  P,.  (/,-",.)•< 

6'  E  I, 
v=l 

M,  e,.  (~>  lr-ev  +  $r  rr  (3  /,.-^  e,.) 


-  ^'P,  (/,.-«,)'v-^  (/,-r,,)  ^  />,  (c.-a,)^      .    .    .    (40) 
From  (10), 


APPENDIX.  93 


But,  since  AY—      1,  this  becomes 


(10a) 


Substituting  (10a)  in  (40),  and  solving  for  f0  we  obtain 
hf+.   -h,.  1          (  , 


-f  I  Pr  ar  ft.— ar)  (#  /,.— ar) 
't?=jf 

-  ^'  A,  |  2  M;.  *  (5^,-^^-  -^-) 


i  1  yf"  9        /    fo  ^    *'V       \ 

|     7lf,.+  /    ^    (o  —      — — ) 

+    4    (•>  —     -^L)     "     Pr    (Ir-a,.)—?  Pr    (e~drY 

-3  (/,— er)  I  Pr  (e—  a,.)2 1 (41) 

Since  O,.=AY  /,,  a,  (l—ar)   (2  lr—ar)=l?  (2  kr—3  ^+K), 
/r — a,.— 7(.  (1 — AY),  and  (41)  reduces  to 

tr—         —7—  ...     I   -  r       ,--N     ,-H      r 


-  ->'  (/,— e,.)l' P,  (e—  a*)*  I    .    .        .    .  (42) 
Returning  to  (24), 

'',":   =^  4rf-^4^-:  .020 


94  APPENDIX. 

Substituting  (32)  and  (35)  in  (30),  and  the  result  in  (24), 
it  becomes 

=  tr  +  ^~-r  \  2  Mr  x~r+Sr  *t-SPr  (*,-a,)> 
2  E  I,    ( 

v=l 


3/r  «v-f-Sr  <£—-!'  Pr  (<v_  a,)*      .    .    .    .(43) 

Making  x,.=lri  then  -^|-         /,.  ,,,  c^ch  and/,.     ),,   and 

substituting  for  S,.  its  value  from  (10),  t,  from  (42),  and 
A1,  I,  for  rr,  in  (43),  it  becomes 


I  -  }  2  Mr  1,+M,.  ,  /, 

0    J^j  _/ . 


-  S    Lr\;>Mf  e,  (3  1—3  er 
';  E  I,  /,        ,       l 


,.l;  (2  fcr-5« 


v=l 


+9  J  P,  (l—kr)  £  1..—3  lt.  2  P;.  (rr— a,)*  \    .    .    .    .  (44) 

) 

Which  reduces  to 

hr+l—'hr       +         _!_ |    M    ^^   M          ^s  p     ^    ^—ft)     1 


APPENDIX.  95 


If  we  were  to  suppose  loads  in  the  v — 1th  span  at  distances 
<7,._;--Av_/  /,._/  from  the  left  support  r-—l,  we  would  find  in 
a  similar  manner,  or  by  decreasing  the  subscripts  of  (45).  by 
unity, 

k-h,._.  1 


•    •    -(46) 


Equating  (42)  and  (46),  we  obtain 

^~hf  +  -2  Mr  lr—Mr+l  l—Z  P,  g  (^  fcr 


--  ^     A.  ^  f  •'  -  ^r  +  4-  }  2  M-  ~  -     A.  4 

'     • 


- 

'  ;•  y 

v=l 


?Prlr(i-kr)  4  i'     A, 


APPENDIX. 

r      / 
r  l^+S  P,_,  /;_,  (fc^-#_;)  -f    i-        "7 


>S    Pr-,    U    (1—kr-,)      -   -    ^       A /-/-''•      ?   I*,-,    (c,-ar_,Y 

V=lr-< 

V-=l 

v=lr_, 
Let 
v=l 

?    A,  f*-  ^r-  +  4]=F' ^48) 

v  ',•  ',-    y 

t;=l 


-  (50) 
(51) 
SPf  (er-aiY-Hr.  .(52) 


v  P  (  P  _  /^  ^3  c?  ^  ^ 

f  ^j~  ?  P,.  (e—cty  t    Hr>   .    .   (53) 


Substituting  (48),  (49),  (50),  (51),  (52)  and  (53)  in  (47), 
transposing  and  reducing,  we  obtain 


APPENDIX.  97 

or  Or-*   ^=|-^-  f  2  Mr  /,  ",_,-{-  1/,^  /,  C,-hS  Pr  1;  (2  kr—3  ^ 

,'!'  Prlr(l—kr)  Or_, 


;_t  M,_,  Br+FZ-t  2  Mr  9r+F^ 
±4  0=0    ,  .    .  (54) 

Which  reduces  to 


r+1  ()  or_,=    -  or  or_t 


I        (,._/  t,. 

-r  p,  //  (^  ^r-^  «+«)  O,._—F;  i  pr  ir  (i—kr}  or_s 


(55) 
Let 

F)  ......    ...    (56) 


-?Pr  l;(2k,.-<  3  ](*+](?)  or_t=A,  ..........   (57) 

-  I  Pr  l~  (k—ty  (>r+!=B,  ........    .......  (58) 

-F;  i1  pr  i,.  (i—icr)  0,-,+Hr  or_,  =x;  .  .  |  .  .  .  .  (59) 

-2  F;_,  Z  P,_;  U  (i—  tU)  ^-^--/  ^r=-X^/  )  ~    •'  •     (59a) 
A  ...................    (60) 


=^f  .    .  .  ................  (62) 

Then  we  can  write 

Ml._l^_l+2Mrft^Mr+.l^f=Ar^rBr_i~}-Yr^X1  .....  (63) 

Which  is  the  general  form  of  the  Theorem  of  three  moments. 
It  expresses  the  relation  between  the  bending  moments  over 
three  consecutive  supports  in  terms  of  the  loads,  spans,  Et 
Tand  /i.  An  equation  of  the  form  of  (63)  can  be  written 
for  every  support,  and,  as,  if  the  girder  rests  on  the  supports, 
the  moment  at  the  firgt  and  last  support  is  zero,  we  shall 


98  APPENDIX. 

have  as  many  equations  as  there  are  unknown  quantities  or 
bending  moments,  and  hence  we  can  determine  their  values. 

Let  there  be  s  spans  in  the  continuous  girder  represented 
in  Fig.  1,  and  let  the  /'"'  span  alone  be  loaded,  then  by  the 
three  moment  theorem,  the  equation  for  each  support  is  as 
follows:  . 

.!  M,    ^    4  ^f    &  —0  i 

M,    &     \  2  ^r!    &    4  M4    &  =o 

XX  XX  XX 

M,.-t  '&_  4-  *  Mf    K    4  M*,  K      K    43  A;  \  ... 

Mr     0r     4-  2  Mf+l  ^  4-  Mr+2  ^=Br 

XX  XX  XXX 


Multiplying  the  first  equation  of  (64)  by  #,,  the  second 
by  c;i}  the  i^*  by  r  ,,  etc  ,  we  obtain,  after  reduction, 


;  cs_;H-^  rs  /SJJ^r,  <?rH  X  O--^,  O  B,  r,.^  .    .  (65  ) 


Now,  supposing  it  is  desired  to  determine  the  value  of 
J!C,  it  is  only  necessary  to  impose  such  conditions  upon  the 
multiplier  c  that  all  terms  shall  reduce  to  zero,  excepting 
those  containing  JfcT,.  Evidently,  the  co-efficients  must 
separately  equal  zero,  or 

£-<y''£s  4  c;!  fa=o          *\ 
c,  p"3    4   £  c,  fa   4    c4  £,=0 
rs  ^    4   ^  c.  fc    -    rj.ft=M9  -    ........    (66) 

rm_,  fi'in_,   4-  2  CM  [i'm  4  fl)l+/  j5IH=0  J 
And,  for  Jjf,,  we  have,  at  once, 

r)f,^rCf-:B,c^  67 

'  '  '  ' 


APPENDIX.  99 


In  a  similar  manner,  multiplying  the  last  equation  of 
(64)  by  ds,  the  last  but  one  by  d;i,  and  so  on,  we  obtain 

•;  d.,  fc  (I.  &L<=0  "| 

.?    -  2         •         = 

.....    (68) 


,,,_,    _,,,r, 
And 


,       _    (r+d^+Ard^+Brd^, 

(2&rf.+  cU.,y  =     -d.+lfi 

From  (65),  we  see  that  M:i=        "-^-  MSJ  from  (66),  c3= 

2    £}    ft' 

—  ;  hence,  assuming  cf=0  and  c,=l,  we  have  M3=cs 

Pa 

^-  Mgj  and  in  a  similar  manner  we  find  that  M4=c4    '^'^ 

Pa  Pa  PS 

3/2,  MB=cB  '*  (  rj,  '  rj,    Mg,  or  in  general  for  any  support  on  the 

Pz  Ps  PA 

left  of  the  loaded  span  or  spans. 

•  Jm-t      jir  /rn\ 

-7  ---    Ms     .............   (/Uj 


Pa   tj:;  •     •      »i-i 

In  a  like  manner,  we  obtain  for  any  support  on  the  right 
of  the  loaded  span  or  spans, 

m>r  v 

IT  fl  .    P«-J    t*--'    •        •    Pm       I  T  /ni\ 

Mm=ds_m+2  -  -  ^—      -—  -  Mfl  ............  (71) 

/-»«_/    Ps—  2    •       '    lj,i, 

Substituting  (69)  in  (70),  and  (67)  in  (71),  we  obtain 


/7O\ 

/;  7?rtt  \          ~+>  '  ~+>  (72) 


•  •  •  <73> 


In  (72),  as  in  must  always  be  less  than  -/*-r  i,  r  can  have 
values  from  .s  to  in. 


100  APPENDIX. 

In  (73"),  as  m  must  always  be  greater  than  f,  /'  can  have 
values  from  1  to  m  —  1. 

Hence,  adding  (72)  and  (73),  and  substituting  X'  -f  X,_, 
for  Xn  we  have 


From  (J),  we  can  obtain  the  bending  moment  over  any 
support  m  of  a  continuous  girder  of  any  number  of  spans  s, 
of  any  lengths  as  lt,  I,  .  .  .  18,  supports  at  any  levels,  the 
moment  of  inertia  I  constant  or  variable,  the  modulus  of 
elasticity  JE  being  constant,  and  the  loads  being  placed  at 
pleasure. 

Note  that  £'</?„_„  &<£*-/,  /C/^*,  and  /5fj_/>/?w. 


TABLE  I. 


101 


*4 

t-f 

a 

/l"7~ 

A-  -A" 

Hr 

A-  -A;* 

oe-o 

059  784  COO 

.940 

.120 

118  272  000 

.880 

.001 

COO  999  999 

.999 

'  61 

060  773  019 

39 

21 

119  228  439 

79 

2 

001  999  992 

8 

62 

061  761  672 

38 

22 

120  184  152 

78 

3 

002  999  973 

7 

63 

062  749  953 

37 

23 

121  139  133 

77 

4 

003  999  936 

6 

64 

063  737  856 

36 

24 

122  093  376 

76 

5 

004  999  875 

5 

65 

064  725  375 

35 

25 

123  046  875 

75 

6 

005  999  784 

4 

66 

065  712  504 

34 

26 

123  999  624 

74 

006  999  657 

3 

67 

066  699  237 

33 

27 

124  951  617 

73 

8 

007  999  488 

2 

68 

067  685  568 

32 

28 

125  902  848 

72 

9 

COS  999  271 

1 

69 

068  671  491 

31 

29 

126  853  311 

71 

.010 

009  9S9  000 

.990 

.070 

069  657  000 

.930 

.130 

127  803  000 

.870 

11 

010  998  669 

89 

71 

070  642  089 

29 

31 

128  751  909 

69 

12 

Oil  998  272 

88 

72 

071  626  752 

28 

32 

129  700  032 

68 

13 

012  997  803 

87 

73 

072  610  983 

27 

33 

130  647  363 

67 

14 

013  997  256 

86 

74 

073  594  776 

26 

34 

131  593  896 

66 

15 

014  996  625 

85 

75 

074  578  125 

25 

35 

132  539  625 

65 

16 

015  995  904 

84 

76 

075  561  024 

24 

36 

133  484  544 

64 

17 

016  995  087 

83 

77 

076  543  467 

23 

37 

134  428  647 

63 

18 

017  994  168 

82 

78 

077  525  448 

22 

38 

135  371  928 

62 

19 

018  993  141 

81 

79 

078  506  961 

21 

39 

136  314  381 

61 

.020 

019  992  000 

.980 

.080 

079  488  000 

.920 

.140 

187  256  000 

.860 

21 

020  990  739 

79 

81 

080  468  559 

19 

41 

138  196  779 

59 

22 

021  989  352 

78 

82 

081  448  632 

18 

42 

139  136  712 

58 

23 

022  987  833 

77 

83 

082  428  213 

17 

43 

140  075  793 

57 

24 

023  986  176 

76 

84 

083  407  296 

16 

44 

141  014  016 

56 

25 

024  984  375 

75 

85 

084  385  875 

15 

45 

141  951  375 

55 

26 

025  982  424 

74 

86 

085  363  944 

14 

46 

142  887  864 

54 

27 

026  980  317 

73 

87 

086  341  497 

13 

47 

143  823  477 

53 

28 

027  978  048 

72 

88 

087  318  528 

12 

48 

144  758  208 

52 

29 

028  975  611 

71 

89 

088  295  031 

11 

49 

145  692  051 

51 

.030 

029  973  000 

.970 

.C90 

089  271  000 

.910 

.150 

146  625  000 

.850 

31 

030  970  209 

69 

91 

090  246  429 

9 

51 

147  557  049 

49 

32 

031  967  232 

68 

92 

091  221  312 

8 

52 

148  488  192 

48 

33 

032  964  063 

67 

93 

092  195  643 

7 

53 

149  418  423 

47 

34 

033  960  696 

66 

94 

093  169  416 

6 

54 

150  347  736 

46 

35 

034  957  125 

65 

95 

094  142  625 

5 

55 

151  276  125 

45 

36 

035  953  344 

64 

96 

095  115  264 

4 

56 

152  208  584 

44 

37 

0-)6  949  347 

63 

97 

096  087  327 

3 

57 

153  130  107 

43 

38 

037  945  128 

62 

98 

097  058  808 

2 

58 

154  055  688 

42 

39 

038  940  681 

61 

99 

098  029  701 

1 

59 

154  9KO  321 

41 

.040 

039  936  000 

.960 

.100 

099  000  000 

900 

.160 

155  904  COO 

.810 

41 

040  931  079 

59 

1 

099  969  699 

899 

61 

156  826  719 

39 

42 

041  925  912 

58 

2 

100  938  792 

98 

62 

157  748  472 

38 

43 

042  920  493 

57 

3 

101  907  273 

97 

63 

158  669  253 

37 

44 

043  914  816 

56 

4 

102  875  136 

96 

64 

159  589  056 

36 

45 

044  908  875 

55 

5 

103  842  375 

95 

65 

160  507  875 

35 

46 

045  902  664 

54 

6 

104  808  984 

94 

66 

161  425  704 

34 

47 

046  896  177 

53 

7 

105  774  957 

93 

67 

162  342  537 

33 

48 

047  889  408 

52 

8 

106  740  288 

92 

68 

163  258  368 

32 

49 

048  882  351 

51 

9 

107  704  971 

91 

69 

164  173  191 

31 

.050 

049  875  000 

.950 

.110 

108  669  000 

.890 

.170 

165  087  000 

.830 

51 

050  867  349 

49 

11 

109  632  369 

89 

71 

165  999  789 

29 

52 

051  859  392 

48 

12 

110  595  072 

88 

72 

166  911  552 

28 

53 

052  851  123 

47 

13 

111  557  103 

87 

73 

167  822  283 

27 

54 

053  842  536 

46 

14 

112  518  456 

74 

168  731  976 

26 

55 

054  8  {3  625 

45 

15 

113  479  125 

•85 

75 

169  640  625 

25 

56 

055  824  384 

44 

16 

114  439  104 

84 

76 

170  548  224 

24 

57 

056  814  807 

43 

17 

115  398  387 

83 

77 

171  454  767 

23 

58 

057  804  888 

42 

18 

116  356  968 

82 

78 

172  360  248 

22 

59 

058  794  621 

41 

19 

117  314  841 

81 

79 

173  264  661 

21 

>k_w+l* 

a 

*i*-lM* 

i.  « 

^  k-'i  A"+A"V 

,   a 

/-=  

I 

/ 

"   i 

102 


TABLE  I.— Continued. 


a 
' 

k-k' 

1 

1    n  ! 

fc=-7-   I:  k:: 

i 

fc—fc8 

\ 

.180 

174  108  000 

.82) 

.210 

220  176  000 

.760 

.300 

273  000  000 

.700 

81 

175  070  259 

19 

41 

227  002  479 

59 

1 

273  729  099 

699 

82 

175  971  432 

18 

42 

227  827  512 

58 

2 

274  456  392 

MS 

83 

176  871  513 

17 

43 

228  651  093 

57 

3 

275  181  873 

97 

84 

177  770  496 

16 

44 

2i9  473  216 

56 

4 

275  1)05  536 

06 

85 

178  668  375 

15 

45 

230  293  875 

55 

5 

276  627  375 

86 

179  565  144 

14 

46 

231  113  061 

54 

6 

277  347  384 

01 

87 

180  460  797 

13 

47 

231  930  777 

68 

7 

278  065  557 

93 

88 

181  355  328 

12 

48 

•J.T2  747  008 

52 

8 

278  781  888 

92 

89 

182  248  731 

11 

49 

233  561  751 

51 

9 

279  496  371 

91 

.190 

183  141  000 

.81CU 

.250 

234  375  000 

.750 

.310 

280  209  000 

.600 

91 

184  032  129 

9 

51 

235  186  749 

18 

11 

280  919  769 

89 

92 

184  922  112 

8 

52 

235  9!)«  !)!I2 

48 

12 

281  628  672 

88 

93 

185  810  943 

7 

J-3 

236  805  723 

47 

13 

282  335  703 

87 

94 

186  698  616 

6 

54 

237  612  936 

46 

14 

283  010  856 

86 

95 

187  585  125 

5 

55 

238  418  6.5 

45 

15 

283  744  12.') 

85 

96 

18*  470  464 

4 

56 

239  222  784 

44 

16 

2<si  44  ')  504 

84 

97 

189  354  627 

3 

57 

240  025  407 

43 

17 

285  144  987 

83 

98 

190  237  608 

2 

58 

240  826  488 

42 

18 

235  842  :>(iS 

82 

9} 

191  119  401 

1 

59 

241  626  021 

41 

19 

286  538  241 

81 

.200 

192  000  000 

.8rO 

.WO 

$•42  424  O'X) 

.740 

.320 

287  232  000 

.680 

1 

192  879  399 

.799 

61 

243  220  419 

39 

21 

287  923  88!) 

79 

2 

193  757  592 

98 

62 

244  015  272 

38 

22 

288  613  752 

78 

3 

191  634  573 

97 

63 

244  808  553 

37 

23 

289  301  733 

77 

4 

195  510  336 

96 

64 

245  600  256 

88 

24 

289  987  776 

76 

5 

196  384  875 

95 

65 

246  390  375 

35 

25 

290  671  875 

75 

6 

197  258  184 

94 

66 

247  178  904 

34 

26 

291  354  024 

74 

7 

198  130  257 

93 

67 

247  9a5  837 

33 

27 

292  034  217 

73 

8 

199  001  088 

92 

68 

248  751  168 

32 

28 

292  712  448 

72 

9 

199  870  671 

91 

69 

249  534  891 

31 

29 

293  388  711 

71 

.210 

200  739  000 

.790 

.270 

250  317  000 

.7450 

.aso 

291  063  OfO 

.670 

11 

201  606  069 

89 

71 

251  097  489 

21) 

31 

294  735  309 

69 

12 

202  471  872 

88 

72 

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73 

311  110  083 

27 

33 

254  990  463 

67 

93 

180  878  043 

7 

74 

310  315  176 

26 

34 

253  906  296 

66 

94 

179  483  016 

6 

75 

309  515  625 

25 

35 

252  817  125 

65 

95 

178  082  625 

5 

76 

308  711  424 

24 

36 

251  722  944 

64 

96 

176  676  864 

4 

77 

307  902  567 

23 

37 

250  623  747 

63 

97 

175  265  727 

3 

78 

1307  089  048 

22 

38 

249  519  528 

62 

98 

173  849  208 

2 

79 

30S  270  861 

21 

39 

218  410  281 

61 

99 

172  427  301 

1 

a 

g  /•  ;  i>*+l>3  7.   a 

a 

-.  -*  k~~~r 

I 

t 

106 


TABLE  I.— Continued. 


I.  a 

"  / 

A  ~   A 

-T 

ik—jb8 

Hr 

A--  -/r 

.900 

171  000  000 

.100 

.940 

109  416  000 

60 

.980 

03*  808  000 

20 

1 

169  5t7  299 

99 

41 

107  762  379 

59 

81 

036  923  859 

19 

2 

168  129  192 

98 

42 

106  103  112 

58 

82 

035  033  H32 

18 

3 

166  685  673 

•   97 

43 

104  43S  193 

57 

83 

033  137  913 

17 

4 

165  236  736 

96 

41 

102  7(57  till! 

66 

84 

031  236  098 

16 

5 

1(53  782  375 

95 

45 

101  091  :!7"> 

55 

86 

029  328  375 

15 

6 

162  322  584 

94 

46 

099  409  464 

54 

86 

027  414  744 

14 

160  857  357 

93 

47 

097  721  877 

53 

87 

025  495  197 

13 

8 

159  386  688 

92 

48 

096  028  608 

52 

88 

023  5(59  728 

12 

9 

157  910  571 

91 

49 

094  329  651 

51 

89 

(2L  038  331 

11 

.910 

156  429  000 

90 

.950 

092  (525  COO 

50 

.990 

019  701  000 

10 

11 

154  941  969 

89 

51 

090  914  649 

49 

91 

017  757  72!) 

9 

12 

153  449  472 

88 

52 

089  198  592 

48 

92 

015  808  512 

8 

18 

151  951  -503 

87 

53 

087  47U  823 

47 

93 

013  8.-,:!  ;M3 

7 

14 

150  448  056 

86 

54 

085  749  :m 

16 

94 

Oil  892  21  (i 

(i 

15 

148  939  125 

85 

55 

084  016  125 

45 

95 

009  925  125 

6 

16 

147  424  7(14 

84 

56 

082  277  184 

44 

96 

007  952  064 

4 

17 

145  904  787 

83 

57 

(  s  i  r,32  507 

18 

97 

005  973  027 

3 

18 

144  379  368 

82 

58 

078  782  088 

42 

98 

003  998  (X  8 

2 

19 

142  848  441 

81 

59 

077  025  921 

41 

99 

TOl  997  001 

.001 

.920 

141  312  000 

80 

.96) 

075  264  000 

40 

n 

21 

139  770  039 

79 

61 

073  49(5  819 

39 

2  k-3  Ir  /,•'• 

fc=  — 

22 

138  222  552 

78 

62 

071  722  872 

38 

/ 

23 

136  (569  533 

77 

63 

069  943  653 

:!7 

24 

13T>  110  976 

7(5 

64 

068  158  (-..Hi 

36 

26 

133  S46  875 

:r> 

65 

(KK>  3(57  875 

35 

26 

131  977  224 

74 

66 

0(51  571  304 

34 

27 

130  402  017 

73 

67 

062  7(58  937 

33 

28 

128  821  248 

72 

(18 

060  960  7(58 

32 

29 

127  234  911 

71 

69 

OE9  146  791 

81 

.930 

125  643  000 

70~ 

~970 

057  327  000 

30 

31 

124  045  f>()9 

69 

71 

055  501  38!  » 

29 

32 

122  442  432 

68 

72 

053  669  !»52 

28 

33 

120  833  763 

67 

73 

051  S32  liS3 

-'7 

34 

119  219  496 

66 

74 

049  !IH!t  :>7(i 

26 

35 

117  599  625 

65 

75 

048  140  625 

25 

36 

115  974  144 

64 

76 

046  285  824 

24 

37 

114  343  047 

63 

77 

044  425  Ki7 

23 

38 

112  706  328 

62 

78 

C42  558  648 

22 

39 

111  (63  981 

61 

79 

040  686  261 

21 

2  k-3  k^k3 

Hr 

2  k-i  k^+k1 

k~  a 
k~  I 

REFERENCES. 


The  Britannia  and  Conway  Tubular  Bridge,  with  general  inquiries  on 
Beams  and  on  the  properties  of  Materials  used  in  Construction. — Edwin  Clark, 
London,  1850. 

Calcul  d'une  poutr£  elastique  reposant  librement  sur  des  appuis  inegalement 
especes.— Clapeyron.  1857. 

Beitrag  zur  Theorie  der  Holz  und  Eisen  constructionen.— Mohr,  1860. 

Beitrage  zur  Theorie  continuirlicher  Bruckeutrager.—  Winkler,  Civil  Iug<5- 
nieur,  1862. 

Die  Lehre  von  der  Elasticetaet  und  Festigkeit,—  Winkler,  18(i7. 

-Vortrage  viber  Briickenbau.—  Winkler,  1875. 

-Theorie  der  Briicken  Aeussere  Kraf  tegerader  Trager.—  Winkler,  Wien,  1875. 

Course  de  Mechanique  Appliquee.    Ill  Partie.— Bresse,  Paris,  1865. 

-Die  graphische  Statik.— Culmann,  1866. 

A  Handy  Book  for  the  Calculation  of  Strains,  &c.—Humber,  New  York,  1860. 

Calcul  des  Fonts  mgtalliques  a  poutres  droites  et  continus.— DeMoudesir, 
Pans.  1873. 

=:  Allgemeine  Theorie  und  Berechnung  der  continuirlichen  und  einfachen 
Trager.—  Weyrauch,  Leipzig,  1873. 

The  Elements  of  Graphical  Statics.— DuBois,  New  York,  1875. 

Graphical  Method  for  the  Analysis  of  Bridge  Trusses.— Greene,  New  York, 
1875. 
Continuous  Revolving  Draw  Bridges.— Herschel,  Boston,  1875. 

Der  bau  der  Briickentrager  mit  Besouderer  Biicksicht  auf  eisen-construc- 
tiouen,  pp.  161-231.— Laissle-Schiibler,  Stuttgart,  1876. 

The  Theor/ and  Calculation  of  Continuous  Bridges.— Merriman,  New  York, 
1876. 

Practical  Treatise  on  the  Properties  of  Continuous  Bridges.— Bender,  New 
York,  1876. 

Elementary  Theory  and  Calculation  of  Iron  Bridges  and  Roofs.— Bitter— 
(Hankey)-  London,  1879. 

The  Strains  in  Framed  Structures  —DuBois,  New  York,  1888. 


108 


REFERENCES. 


Annales  des  Fonts  et  Chaussces. 


1864—  Paper 

75—  pp.  141-213  

Collignon. 

1866—      " 

126—  pp.  310-408  

Renaiidot. 

1866—      " 

132—  pp.    53-175  

Albaret. 

1867—      " 

158-^pp.    27-115  

Regnauld. 

1871—      " 

3—  pp.    44-51  

Pierre. 

•-••1871-      " 

20—  pp.  170-274  

Des   Orgeries. 

1872—      " 

25—  pp.  189-220 

Poulet. 

1S74—      " 

15—  pp.  327-391  

Choron. 

1882—      " 

9—  pp.  141-218  

Huleu'icz. 

1884—      " 

44-  pp.  101-197  

HulHvicz. 

188-5—      " 

72—  pp.  267-351  

Guillaume. 

1885— 

86—  pp.  613-726  

Hausserct. 

1886-      " 

40—  pp.      5-39  

Collignon. 

1886-      " 

11—  pp.  304-357      

Leygue. 

Van  Nostrand's  Engineering  Magazine. 


*1874— pp.  55i_.V)7  . 

1876— pp.     65-  73  .    . 

*1877— pp.  481-490  .    . 

1878— pp.  55:{-5f)0  .    . 

1881-pp.    49-  59  .    . 

1885— pp.  137-144  .    . 


Eddy. 

Merriman. 

Eddy. 

Hiidyins. 

Kidder.\ 

Berfsford. 


••'•  Variable  Moment  of  Inertia. 


INDEX  OF  EQUATIONS. 


TT""  . 

->  1  ;       • 

A.  .    .    . 
B   .    . 
C   .    .    . 
D  .    .    . 
E  .    .    . 
E,  .    .    . 

PAGK. 

7 
12 
.    ,14 
15 
16 

10a    

PAGE. 

5 

14-19 

6 

20-21 

.    .    7 
.    .  86 
87 

22a-22d    
22c-22m    .    . 

16 
17 
17 
11 

22-26    
27-28    

.    .  88 
89 

29-31     
32-36 

.    .  90 
.    .  81 
.    .  92 
93 

b    .    .    . 
c     .     .     . 

d  .  .  . 

e     .    .    . 
f 

11 
11 
11 
12 
9 

37-40    
41-42    ..... 

43-44    
45a 

.    .  94 
.    .  17 
95 

45-46 

o 

10 

47-53    
54-63    .    . 

.    .  96 
97 

h     .    .    . 

i     .•   .    . 
i.,    .    .    . 
L    .    -    . 
i 

9 
8 
12 
.......  ]4 

8 

59a    

97 

64-67    
68-73    

74-77 

.    .  98 
.    .  99 
.    .    8 
.    .    9 
10 

j>    •    •    • 

L 

12 
14 
.  12 
12 
15 

78-81    
82-85    

86-89 

.    .  13 
14 

90-92 

93 

15 

n    .    .    .    . 

0      .      .      .      . 

12 

94-102  
103-110    ...... 
111-117    
118-124 

.    .  16 
.    .  19 
.    .  20 
21 

12 
12 
12 

12 

125-134    
135-141     
142-152 

.    .  22 
.    .  23 
24 

'D» 

...                  14 

r     .    .    .    . 

T  ,      .      .      .      . 

1-7 

12 
12 
14 
4 

153-162    ." 

25 

163-173    .... 

26 

174-178    

27 

8-13  .    .    . 

5 

179-188    

.    .  28 

110 


INDEX  OF  EQUATIONS. 


189-196 


.  21)     262-2(8  .  39 


197-206  .'.......:'»() 

207-217    31 

218-223 :\-2 

224-233    33 

234 34 

235-238 35 

239-2-12 :-',(> 

243-251 37 

252-261  .  38 


266-275    40 

276-286 41 

28  -291  .  42 


292-300 
301-30(5 
307-311 
312-313 
314-316 


43 
44 

45 
46 
47 
48 


SUMMARY. 


PAGE. 

Nomenclature 1-2 

I. 
GENERAL  RELATIONS. 

Conditions  of  equilibrium — (I),  (II)  and  (III)  ...  4 
Moment  equations  for  single  concentrated  load — (1) 

and  (2) 4 

Moment  equations  for  concentrated  loads — (8)  and  (9)  5 
Moment  equations  for  partial  uniform  loads— (13),  (14) 

and  (15)  5-6 

Moment  equations  for  uniform  load  over  all — (18)  and 

(19)   •    • 6 

MOMENTS. 

General  equations  for  the  moment  over  any  support ;  the  modulus 

of  elasticity,  E,  alone,  being  considered  constant. 
Moment  equation  ( A ) 7 

Concentrated  loads—- 

Values  of  A,  and  Br—(i)  and  (j) 

Values  of  X;,  X^t  and  Hr—  (A),  (h)  and  (/)  ....  9 
Value  of  H;—(g)  ....  10 

Partial  uniform  loads  — 

Values  of  A,,  and  Br—  (74)  and  (76) 

Values  of  A7  and  AZ/— (78)  and  (80) 9 

Value  of  #,— (f),  (82)  and  (83) 9-10 

Value  of  H/—  (g),  (82)  and  (83)      10-11 

Uniform  load  over  all— 

Values  of  Ar  and  B,.— (75)  and  (77) 

Values  of  A7  and  X;_,—  (79)  and  (81) 9 

Value  of  Jfy.— (/),  (84)  and  (85) 9-10 

•Value  of  If/— (g),  (84)  and  (85) 10-11 


112  SUMMARY. 

For  all  loads  — 

Values  of^.,  F/,  F,"  and  ,—(6),  (c),  (rf)  and  («)  .  .  11 
Values  of  ftr ,  ,V,  £',  }',",,  e*  and  d,,—  (m),  (n),  (o),  (/), 

(e),  (p)  and  (r) 12 

The  modulus  of  elasticity,  E,  and  tlie  moment  of  inertia,  I,  being 

constant. 

Moment  equation  (A,) 12 

Concentrated  loads,  values  of  Af  and  B,. — ( i, )  and  ( j, )  .  13 
Partial  uniform  loads,  values  of  ^1,  and  B, — (86)  and  (88)  13 
Uniform  load  over  all,  values  of  A,,  and  B, — (87)  and  (89)  13 

For  all  loads- 
Values  of  c,,,,  d,H  and  Y,—(p,),  (rt)  and  (/,) 13 

The  modulus  of  elasticity,  E,  the  moment  of  inertia,  I,  and  the 
length  of  the  spans  being  constant. 

Moment  equation  (Ax) 14 

Concentrated  loads,  values  oi  A,  and  B, — (L)  and  (,/,)  .  14 
Partial  uniform  loads,  values  of  A,,  and  B,.  — (90)  and  (92)  14 
Uniform  load  over  all,  values  of  A,,  and  B, — (91)  and  (93)  14-15 

For  all  loads- 
Values  of  cw,rfw  and  Yr—(p,),(rs)  and  (/,) 14-15 

GENERAL  EQUATIONS  FOR  SHEAR. 

General  equation  (B),  value  of  S, 15 

General  equation  (C),  value  of  &,! 16 

Concentrated  loads,  values  of  Q;.and  Q/— (94)  and  (95)  16 

Partial  uniform  loads,  values  of  Q,.  and  QJ — (96)  and  (97)  16 

Uniform  load  overall,  values  of  Q,and  Q/—  (98)  and  (99)  1(5 

GENERAL  EQUATIONS  FOR  INTERMEDIATE  BENDING  MOMENTS. 

General  equation  (D) 16 

Concentrated  loads,  value  of  Lr — (100) 5  and  16 

Partial  uniform  loads,  value  of  L, — (101) 6  and  16 

Uniform  load  over  all,  value  of  L,. — (102) 7  and  16 

GENERAL    EQUATIONS    FOR    DEFLECTION. 

E,  alone,  constant. 

General  equation  (£"),  value  of  y, 17 

Value  of  tr+i— (45) 17 

E  and  I  constant. 

General  equation  (£";),  value  of  y, ..-.-.  17 

Value  of  tr+,— (45a) 17 


SUMMARY.  113 

II. 
SUPPORTED  GIRDERS. 

GIRDER  RESTING  UPON  T\VO  SUPPORTS. 

(a)     JE,  alone,  constant. 
Values  of  M,  and  M— (103)  .  19 

Concentrated  loads — 

Values  or  >S  and  £'— (106)  and  (107) 19 

Value  of  J/,'—  (113)      20 

Value  of  y,— (117)'  20 

Partial  uniform  loads- 
Values  of  S,  and  £'— (108)  and  (109) 19 

Value  of  M;— (114) 20 

Value  of  y,— (117)  ...  See  (12),  p.  18  20 

Uniform  load  over  all — 

Values  of  S,  and  &'—  (110)  and  (111) 19-20 

Value  of  MJ— (115) 20 

Value  of  M—  (116) 20 

Value  of  ?/,— (117)  -        -   See  (12),  p.  18  .  20 

(6)     IE  and  I  constant. 
Value  of  y, — (119)    .    .    (concentrated  loads) 21 

A  BEAM  CONTINUOUS  OVER  THREE  SUPPORTS. 

(a)     jEJ,  alone,  constant. 

Value  of  J/;,  M.,  and  M— (120)  and  (121) 21 

Value  of  M— (124) 21 

Concentrated  loads,  values  of  A.>  and  Bt — (135)  and  (136)       23 
Partial    uniform   load,    values   of    A.,    and    B, — (137) 

and  (138) 23 

Uniform   load  over   all,    values  of    A.>  and  B, — (139) 

and  (140) " 23 

For  all  loads— 

Values  of  c,,  c.,  and  £,—(122) 21 

Values  of  </„  d*  and  d—  (123) 21 

Values  of  A7,  A7,  H,  and'  H,'— (125),  (126),   (127) 

and  (128) 22 

Values  of  J,,  z\r,  FJ,  F"  and  Fw— (129),  (130),  (131), 

(132)  and  (133) 22 

Values  of  /V  and  K— (133a)  and  (134) 22 


114  SUMMAHV. 

Shears,  equations  (B)  and  (C) 15-16 

Moments,  intermediate — (D) 16 

Deflection— (E)  and  (45)    .  17 

(6)     E  and  I  constant. 

Values  of  J/;  and  AT--(141) 23 

Value  of  M— (142)  ...  24 

Concentrated  loads — 

Value  of  M— (150) 24 

Values  of  A  and  B—  (144)  and  (145) 24 

Values  of  S,,  £',  S,  and  S/— (153)  to  (157)    ....  25 
Partial   uniform   loads,  values  of   A.,    and    B, — (146) 

and  (147) 24 

Uniform  loa'l  over  all — 

Value  of  M,—(  158) 25 

Values  of  S,,  X/,  ,%  and  ^'—(159)  to  ( 163)     25-26 

For  all  loads- 
Value  of  Y,— (143) 24 

(c)  JEj  I  and  h  constant. 

Same  as  case  (6),  with  Y.,  —  0 26 

(d)  E,  I,  h  and  I  constant. 
Concentrated  loads — 

Value  of  M— (164) 2ii 

Values  of  S,,  £',  S,  and  S'— (165)  to  (169)    .    .  W 

Uniform  load  over  all — 

Value  of  M— (170) 26 

NaluesofX,,  X/',  tf/ancLS,'—  (171)  to  (174) 26-27 

A  BEAM  CONTINUOUS  OVER  FOUR  SUPPORTS. 

(a)     JJEJ,  alone,  constant. 

Values  of  J/,  and  3/,-(175; 27 

Values  of  M,  and  M—(  177)  and  (178) 

See  General  Relations  for  other  equations  ....  27 

(6)     E  and  I  constant. 

Values  of  M,  and  , I/— (179)  and  (180) 28 

See  General  Relations  for  other  equations  . 

(c]     JEJ,  I  and  h  constant. 

Values  of  M,  and  M— (181)  and  (182) 28 

See  General  Relations  for  other  equations 


SCMMAKY.  115 

(  d }      H,  /,  h  <in<l  /  cniixtniit. 
\ allies  of  .17,  and  .17— ( .  1  S3)  and  (184)  .  2S 

rniform  load  over  all- 
Values  of  .17,  and  .U— (185)  and  (18(3)  .    .    .  2S 

rniform  load  over  all  and  spans  equal— 
Values  of  .17,  and  J/:,— (187)  and  (1S8)  .    .  .    .        2S 

Values  of  X;,  X/,  N,,  >',',  X,  and  X,"— (189)  to  (196)     .    .        k2l) 

TIIK    TIIM'ER. 

(  d  )      K,  (done,  ronx((tnt. 

Value  of //.,—( 217) 

Values  of  V.,  and  )>— (210)  and  (211)     ....... 

Values  of  .17,  and  .I/,— (177)  and  (ITS)  .    .    .    .    . 

See  General  Relations  for  other  equations 

( I) }      E  and  I  convlant. 

Values  of  T.  and  Y     (233) 33 

Values  of  .17,  and  .17— (179)  and  (180)    .    .  28 

See  General  Relations  for  other  equation^ 


III. 

BEAMS  WITH   FIXED  ENDS. 

A    MKAM    KIXKI)   AT  ONE  KM)   A  XI)    SI   l'l'(  >RTKI  >   AT  TIIK  OT 


Value  of  Mr- (234) 34 

Concentrated  loads,  value  of  A,, — ( 144) 

Partial  uniform  loads,  value  of  A., — (146; 

Uniform  load  over  all,  value  of  A., — (148)  .    .        ... 
For  all  loads 

Values  of  )'.,,   B'  and    A,— (236)  to  (  129) 35 

Values  of  7'V,  7-;  11,  and  AY— (131)  to  (238) 35 

See  ( General  Relations  for  other  equations 

( ft )      K  « ml  I  roiiNffi lit. 
Value  of  .17,— (240)    .    .    .  36 

Concentrated  loads- 
Value  of  .17,— (241) 36 

Value  of  )',— (236) 35 

Values  of  X,  and  X'     (212)  and  (243)   .  .  36-37 


116  SUMMARY. 

Uniform  load  over  all— 

Value  of  .)/.,— (244)   .    .    .  :'>7 

Value  of  K,— (236)  .    .    .  85 

Values  of  X  and  X'— (245)  and  (24(5)  .  87 

(c)     .H,  1  "/id  It  constant. 

Concentrated  loads — 

Value  of  Mr- (247)  .  37 

Values  of  X  and  X,.'— ( 24s )  and  ( 249 ,  8 , 

("niform  load  over  all- 
Value  of  M;— (250) 37 

Values  of  X,  and  S,'—  (251 )  and  (252)  .  8,-8S 

Value  of  ,!/,—(  253)  . 
Value  of  M«.r.  .I/,— (254)     . 

Load  at  center  of  beam— 

Value  of  .JA— (256) 

Value  of  X,  and  X,.'— (257)  and  (25X>  .  8,s 

Value  of  M  —  (251))  or  (260)  .  :'.S 

\ralue   of  Mas.   :]/,— (261)  .  38 

N'alue  of//,,  if/;,     />,     <>— (2«2)  or  (2r,:V)  .  3<) 

A     1JEA.M     FIXKD    AT     I'.OTII     ENDS. 

(a  )      /£,  nlnin  .  a m slant. 

N'alues  J/,  and  .)/,— (266)  and  (267)     .  40 

Values  of"^'.  A',  A,  &",  ^  and  );-(268)  to  (273)  .    .        40 
See  Generai  Relations  for  other  equations. 

(/;)      E  and  I  constant. 
Values  of  .!/,  and  M:—  (274)  and  (275)  40 

( c )     .E,  I  an d  h  con *ta-nt. 
Values  of  .17,  and  M— (276)  and  (277)  41 

Concentrated  loads — 
Values  of  ,]/,  and  M: .— (27«)  and  (27?h 

Uniform  load  over  all- 
Values  of  .17,  and  : I/.— (280)   . 
\'alues  of  X,  and  X,'— (281) 

Value  of  M—  (282) 41 

N'alue  of  Mas.  J/— (283)  .    .  41 

Value  of?/,— (284) 


SrM.MAKY.  11, 

A  single  load  in  the  center  of  the  beam — 

Values  of  M,  and  M—  (285) 

Values  of  &  and  &'— (286)  .    . 

Value  of  J^— (287)  or  (288)   ...  42 

Value  of  Mas.  #£—(289)  .  42 

Valueof^— (290)  or(291)  .    .    ,  42 


A    I5EAM    FIXED   AT  ONE   END,    AND   I   XST !'!'( )  KTET)    AT  TIIK  OTIIEIl. 

Concentrated  loads — 

Value  of  M— (21)2)  .    .    .  43 

X'alue  of  X,— (292)     ....  43 

Partial  uniform  loads — 

Value  of  M,—( 293) 43 

Value  of  X,— ( 294) 43 

Value  of  M'—  (295)   ,    .  43 

Value  of  ;?/,, — I  constant — (297) 43 

Single  concentrated  load,  value  of  y, — (298) 43 

Load  at  end  of  beam,  value  of  y, — (299) 43 

Uniform  load  over  all,  value  of  ;//, — (300).    .    .  43 

A     I5EAM    OX  TWO  Sl'I'I'OUTS,    AND  ONE   KXD   1   XSI    I'I'oKTED. 

Value  of  M—  (301)  .  44 

Value  of  X,  ,SV  and  X,— (302)  to  (304)  .    .  44 

Values  of  .!/,  and  .I/,— (305)  and  (30(i) 44 

\    P.EA.M   ON  ONE  Sl'J'I'ORT,    HAVIN(i    OXE    END    FIXED,    AXD  THE  OTIIEI! 
i   NSIIM'OKTED. 

(  a  )      12,  afoiWj  constant. 
X'alues  of  .)/,  and  M:I— (309)  and  (307)     .    .    .  4") 

( b )  K  an d  1  constant. 

N'alues  of  M,  and  .]/,-  (310)  and  ( 307  )     .  45 

(c)  H,  I  and  h  cr instant. 

\' ulues  of  .I/,,  and  M— (31 1  )  and  (307)    ...  45 

A    I'.KA.M   ON    T\\  ()  sri'l'ORTS,    IfAXINt.    XKITIIEK    EN  D  SI    I'  I'OUTEI). 

Values  of  M,  and  .17^ — (312)  and  (313)  40 


1 18  SUMMARY. 

IV. 
THE   mi  NT  OK   XKUo   MOMENT. 

Load  on  the  right,  value  of  .»-,. — ('>\\<>) 47 

Load  on  the  left,  value  of  .rr — (317)     4s 

Values  of  .r,  by  (Jraphics L8-49 

T1IK  CO-KFKICIKNTS  <-  AND  </. 
(Iraphical  determination  of  the  values  of  <•  and  <l    .    .  -">0-5i 


APPLICATIONS. 

Kxainple  I.— A  continuous  girder  of  three  spans  upon 
level  supports,  the   two  end  spans  being  equal. 
Determination  of  bending   moments  and  shears       o2 
Values  of  .17,  and  .17..  for  each  load — Table  (H)    ....  08-54 

Values  of  N,  >./,  >',  >',',  >',  and  X/— Table  (/>) -V> 

Values  of  .17,.— Table' (c)  -^ 

Maximum  mom^n'ts        ">* 

Maximum  shear -V.t 

Example  1. — By  graphics 60-ttl 

Kxample  '2. —  A.  continuous  girder  of  four  spans  with 
uniform  loads  and  a  variable  cross  section.  De- 
termination of  moments (;•_!-<;!> 

Kxample  2. — With  a  constant  cross-section 7<»-71 

A  comparison  of  the  results  obtained,  considering  the 

moment  of  inertia  as  variable  and  then  constant  .        71 
Kxample  o. — A  continuous  girder  of  two  spans.     The 
Sabula    Draw   uniformly   loaded,   and   having   a 
variable  cross- section.       Determination  of  the  mo- 
ment over  the  second  support 72-7(> 

Example  o«.     The  same  as  Example  3,  but   with   a 

constant  cross-section 77 

Comparison  of  the  results  of  Example  3  and  Example  3</       77 
Example  4. — The   Sabula   Draw,    with    concentrated 
loads  and  a  variable  cross-section.     Determination 

of  the  moment  over  the  second  support 7<v-*0 

Kxample  4.     By  graphics X0-x:l 


SUMMARY.  119 

APPENDIX. 

Demonstration  of  the  equation  of  the  elastic  line  .  .  .  85-88 
General  expression  for  the  Theorem  of  Three  Moments  .  97 
Demonstration  of  Equation  (A) 89-100 

TABLE  I. 
Values  for  k — 2  F+F  and  fc— 1?  for  all  ratios  -j-     =  k 

from  .001  to  .999,  inclusive 101-106 

REFERENCES. 

Some  references  to  monographs,  periodicals,  &c.,  which 

consider  the  theory  of  the  continuous  girder  .    .  107-108 

INDEX. 
Index  of  Equations      . 109-110 


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